Smart Integration Tricks [closed]

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I am in the last year of my school
and studied integration this year



I have done several Integration techniques like Integration



  1. By substitution

  2. By partial fractions

  3. By parts

  4. Trigo. substitutions


  5. Tangent half angle substitution
    and many other basic methods of integration.

So I wanted to ask about some integration tricks that might prove quite helpful.



Not something advanced which is taught at higher level of studies
But some smart integration tricks at school level only.










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closed as too broad by Paul Frost, amWhy, José Carlos Santos, Adrian Keister, Xander Henderson Sep 4 at 1:40


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • Have a look here math.stackexchange.com/questions/2821112/integral-milking
    – Michael Hoppe
    Sep 3 at 18:35










  • For integration by parts have a look at math.stackexchange.com/questions/20397/…
    – Michael Hoppe
    Sep 3 at 18:39










  • Also related: math.stackexchange.com/questions/70974/…
    – Hans Lundmark
    Sep 3 at 20:22














up vote
9
down vote

favorite
6












I am in the last year of my school
and studied integration this year



I have done several Integration techniques like Integration



  1. By substitution

  2. By partial fractions

  3. By parts

  4. Trigo. substitutions


  5. Tangent half angle substitution
    and many other basic methods of integration.

So I wanted to ask about some integration tricks that might prove quite helpful.



Not something advanced which is taught at higher level of studies
But some smart integration tricks at school level only.










share|cite|improve this question















closed as too broad by Paul Frost, amWhy, José Carlos Santos, Adrian Keister, Xander Henderson Sep 4 at 1:40


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • Have a look here math.stackexchange.com/questions/2821112/integral-milking
    – Michael Hoppe
    Sep 3 at 18:35










  • For integration by parts have a look at math.stackexchange.com/questions/20397/…
    – Michael Hoppe
    Sep 3 at 18:39










  • Also related: math.stackexchange.com/questions/70974/…
    – Hans Lundmark
    Sep 3 at 20:22












up vote
9
down vote

favorite
6









up vote
9
down vote

favorite
6






6





I am in the last year of my school
and studied integration this year



I have done several Integration techniques like Integration



  1. By substitution

  2. By partial fractions

  3. By parts

  4. Trigo. substitutions


  5. Tangent half angle substitution
    and many other basic methods of integration.

So I wanted to ask about some integration tricks that might prove quite helpful.



Not something advanced which is taught at higher level of studies
But some smart integration tricks at school level only.










share|cite|improve this question















I am in the last year of my school
and studied integration this year



I have done several Integration techniques like Integration



  1. By substitution

  2. By partial fractions

  3. By parts

  4. Trigo. substitutions


  5. Tangent half angle substitution
    and many other basic methods of integration.

So I wanted to ask about some integration tricks that might prove quite helpful.



Not something advanced which is taught at higher level of studies
But some smart integration tricks at school level only.







integration definite-integrals indefinite-integrals






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share|cite|improve this question













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share|cite|improve this question








edited Sep 3 at 17:28









GoodDeeds

10.2k21335




10.2k21335










asked Sep 3 at 14:58









JIM

385




385




closed as too broad by Paul Frost, amWhy, José Carlos Santos, Adrian Keister, Xander Henderson Sep 4 at 1:40


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as too broad by Paul Frost, amWhy, José Carlos Santos, Adrian Keister, Xander Henderson Sep 4 at 1:40


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.













  • Have a look here math.stackexchange.com/questions/2821112/integral-milking
    – Michael Hoppe
    Sep 3 at 18:35










  • For integration by parts have a look at math.stackexchange.com/questions/20397/…
    – Michael Hoppe
    Sep 3 at 18:39










  • Also related: math.stackexchange.com/questions/70974/…
    – Hans Lundmark
    Sep 3 at 20:22
















  • Have a look here math.stackexchange.com/questions/2821112/integral-milking
    – Michael Hoppe
    Sep 3 at 18:35










  • For integration by parts have a look at math.stackexchange.com/questions/20397/…
    – Michael Hoppe
    Sep 3 at 18:39










  • Also related: math.stackexchange.com/questions/70974/…
    – Hans Lundmark
    Sep 3 at 20:22















Have a look here math.stackexchange.com/questions/2821112/integral-milking
– Michael Hoppe
Sep 3 at 18:35




Have a look here math.stackexchange.com/questions/2821112/integral-milking
– Michael Hoppe
Sep 3 at 18:35












For integration by parts have a look at math.stackexchange.com/questions/20397/…
– Michael Hoppe
Sep 3 at 18:39




For integration by parts have a look at math.stackexchange.com/questions/20397/…
– Michael Hoppe
Sep 3 at 18:39












Also related: math.stackexchange.com/questions/70974/…
– Hans Lundmark
Sep 3 at 20:22




Also related: math.stackexchange.com/questions/70974/…
– Hans Lundmark
Sep 3 at 20:22










7 Answers
7






active

oldest

votes

















up vote
6
down vote













If you're looking for more practice problems, look for problems from MIT Integration Bees. Some of their problems are kind of easy, but there are quite a few gems.



Also, I highly recommend Paul Nahin's Inside Interesting Integrals. This book is incredibly well-written and a pleasure to read. It isn't exactly what you asked for, because it does eventually go above and beyond high school math. However: the first few chapters don't require much more than high school math. And in the later chapters, Nahin introduces the necessary concepts.



Other sources:




  • Integral Kokeboken (http://folk.ntnu.no/oistes/Diverse/Integral%20Kokeboken.pdf) It is written in a language I don't understand, but there are lots of fun practice problems.


  • Advanced Integration Techniques (http://advancedintegrals.com/advanced-integration-techniques.pdf) I didn't like this as much as Nahin's book, but it covers some of the same things, and maybe it's more your thing.





share|cite|improve this answer






















  • any book that you would recommend for practising integration ?
    – JIM
    Sep 3 at 15:12










  • @JIM Nahin's book has some good practice problems. For a lot of problems. I'm pretty sure Integral Kokeboken has some more elementary stuff.
    – aras
    Sep 3 at 15:15










  • Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
    – JIM
    Sep 3 at 15:21










  • @JIM It's an actual book, so I don't have a link. You can buy it off of amazon (amazon.com/…) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
    – aras
    Sep 3 at 15:25






  • 4




    @JIM No, I'm not going to give you a link to the PDF since I'm not sure about StackExchange's policies. If you want a PDF, google resources for getting pdfs of textbooks. There are piracy websites that are easy enough to find and use.
    – aras
    Sep 3 at 15:29

















up vote
5
down vote













I see that nobody has mentioned a Euler Substitution. So I’ll go ahead and add that here. An Euler Substitution is essentially some kind of algebraic substitution for evaluating integrals of the form$$intmathrm dx, Rleft(x,sqrtax^2+bx+cright)$$



First Substitution of Euler: If $a>0$ then make the substitution$$sqrtax^2+bx+c=pm xsqrt a+t$$where either the positive and negative sign can be chosen.



Second Substitution of Euler: If $c>0$ then let$$sqrtax^2+bx+c=xtpmsqrt c$$Solve for $x$ and differentiate to find what $mathrm dx$ is equal to.$$x=frac pm2tsqrt c-ba-t^2$$



Third Substitution of Euler: If the polynomial inside the square root has real roots $alpha$ and $beta$, then let$$sqrtax^2+bx+c=sqrta(x-alpha)(x-beta)=(x-alpha)t$$Therefore$$x=frac alphabeta-alpha t^2a-t^2$$



Now let’s take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, it’s evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but I’ll use the first substitution because it’s nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$






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  • +1 ,I remember seeing these somewhere, thanks for reminding me!
    – Alvin Lepik
    Sep 3 at 17:06

















up vote
4
down vote













Another neat trick is to add different forms of integrals to obtain a much simpler one.



For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=int_-1^1frac11+f(x),dx$$ then we could replace $x$ by $-x$ giving $$I=-int_1^-1frac11+f(-x),dx=int_-1^1fracf(x)1+f(x),dx$$ and adding gives $$2I=int_-1^1,dx.$$






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    up vote
    3
    down vote













    One very simple example that pops to mind (several of my students have been dumbfounded on first sight with the following example).



    Find the anti derivative (there are many such examples)
    $$int fractt+1dt $$
    You add and subtract $1$ in the numerator and voila.




    Another one. Bearing in mind $int frac11+t^2dt = arctan t + C$. Find the anti derivative, say,
    $$int frac19t^2 -6t +2dt $$
    Separate the square part and integrate with respect to $3t-1$. Simplifies to finding
    $$frac13int frac1(3t-1)^2 +1d(3t-1) $$






    share|cite|improve this answer






















    • some other examples could you quote please ?
      – JIM
      Sep 3 at 15:13










    • @JIM I would have posted a similar example, but you mentioned that you are familiar with "partial fractions", which this kinda boils down too, so I thought this would be nothing new.
      – Cornman
      Sep 3 at 15:14










    • @Cornman in my experience, the less experienced practitioner starts trying to solve the problem using unnecessary force. Something along the lines of hunting down a fly with a cannon
      – Alvin Lepik
      Sep 3 at 15:20










    • @AlvinLepik You are absolutely correct .
      – JIM
      Sep 3 at 15:24

















    up vote
    2
    down vote













    Maybe something like this:



    Calculate $intln(x), dx$ where the trick is to write $int 1cdot ln(x), dx$ and then use partial integration.



    And other one, which you probably already know are integrals of this form:



    $int fracf'(x)f(x), dx=ln(|f(x)|)+c$



    Which finds use in many other integrals, often combined with other methods.






    share|cite|improve this answer



























      up vote
      2
      down vote













      Another different trick suppose you want to calculate



      $$I=int_-infty^inftye^-x²dx$$



      It seems hard integration until you square it and use polar coordinates.
      $x=rsin theta$ and $y=rcos theta$
      $$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$






      share|cite|improve this answer



























        up vote
        2
        down vote













        If you integrate even function $f$ on interval which is symmetric around zero, then you could just integrate $f$ on positive (or negative) part of interval. In other words, if $f(-x)=f(x)$ holds, then



        $$intlimits_-a^af(x);mathrmdx = 2intlimits_0^af(x);mathrmdx$$



        Analogously, integrating odd function $g$ on interval which is symmetric around zero will give you $0$:



        $$intlimits_-a^ag(x);mathrmdx = 0$$



        If function $h$ is periodic, with period $T$, then



        $$intlimits_a^bh(x);mathrmdx = intlimits_a+T^b+Th(x);mathrmdx$$
        and
        $$intlimits_a^a+Th(x);mathrmdx = intlimits_b^b+Th(x);mathrmdx$$
        for every real $a$ and $b$.



        You could easily prove any of these statements.






        share|cite|improve this answer




















        • 1 and 2nd result i know but not able to get 3 and 4 one
          – JIM
          Sep 3 at 16:23










        • Use integration by substitution.
          – Nikola Ubavić
          Sep 3 at 16:27










        • no , I am not asking for the proof but what exactly does this result say ?
          – JIM
          Sep 3 at 16:31










        • @JIM he's saying it's a function which repeats, like $ h(x) = sin(x) $. That has period $ T = pi $, so you can integrate over any different repeated period and get the same result or over any interval of length equal to the period.
          – theREALyumdub
          Sep 3 at 19:28


















        7 Answers
        7






        active

        oldest

        votes








        7 Answers
        7






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        6
        down vote













        If you're looking for more practice problems, look for problems from MIT Integration Bees. Some of their problems are kind of easy, but there are quite a few gems.



        Also, I highly recommend Paul Nahin's Inside Interesting Integrals. This book is incredibly well-written and a pleasure to read. It isn't exactly what you asked for, because it does eventually go above and beyond high school math. However: the first few chapters don't require much more than high school math. And in the later chapters, Nahin introduces the necessary concepts.



        Other sources:




        • Integral Kokeboken (http://folk.ntnu.no/oistes/Diverse/Integral%20Kokeboken.pdf) It is written in a language I don't understand, but there are lots of fun practice problems.


        • Advanced Integration Techniques (http://advancedintegrals.com/advanced-integration-techniques.pdf) I didn't like this as much as Nahin's book, but it covers some of the same things, and maybe it's more your thing.





        share|cite|improve this answer






















        • any book that you would recommend for practising integration ?
          – JIM
          Sep 3 at 15:12










        • @JIM Nahin's book has some good practice problems. For a lot of problems. I'm pretty sure Integral Kokeboken has some more elementary stuff.
          – aras
          Sep 3 at 15:15










        • Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
          – JIM
          Sep 3 at 15:21










        • @JIM It's an actual book, so I don't have a link. You can buy it off of amazon (amazon.com/…) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
          – aras
          Sep 3 at 15:25






        • 4




          @JIM No, I'm not going to give you a link to the PDF since I'm not sure about StackExchange's policies. If you want a PDF, google resources for getting pdfs of textbooks. There are piracy websites that are easy enough to find and use.
          – aras
          Sep 3 at 15:29














        up vote
        6
        down vote













        If you're looking for more practice problems, look for problems from MIT Integration Bees. Some of their problems are kind of easy, but there are quite a few gems.



        Also, I highly recommend Paul Nahin's Inside Interesting Integrals. This book is incredibly well-written and a pleasure to read. It isn't exactly what you asked for, because it does eventually go above and beyond high school math. However: the first few chapters don't require much more than high school math. And in the later chapters, Nahin introduces the necessary concepts.



        Other sources:




        • Integral Kokeboken (http://folk.ntnu.no/oistes/Diverse/Integral%20Kokeboken.pdf) It is written in a language I don't understand, but there are lots of fun practice problems.


        • Advanced Integration Techniques (http://advancedintegrals.com/advanced-integration-techniques.pdf) I didn't like this as much as Nahin's book, but it covers some of the same things, and maybe it's more your thing.





        share|cite|improve this answer






















        • any book that you would recommend for practising integration ?
          – JIM
          Sep 3 at 15:12










        • @JIM Nahin's book has some good practice problems. For a lot of problems. I'm pretty sure Integral Kokeboken has some more elementary stuff.
          – aras
          Sep 3 at 15:15










        • Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
          – JIM
          Sep 3 at 15:21










        • @JIM It's an actual book, so I don't have a link. You can buy it off of amazon (amazon.com/…) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
          – aras
          Sep 3 at 15:25






        • 4




          @JIM No, I'm not going to give you a link to the PDF since I'm not sure about StackExchange's policies. If you want a PDF, google resources for getting pdfs of textbooks. There are piracy websites that are easy enough to find and use.
          – aras
          Sep 3 at 15:29












        up vote
        6
        down vote










        up vote
        6
        down vote









        If you're looking for more practice problems, look for problems from MIT Integration Bees. Some of their problems are kind of easy, but there are quite a few gems.



        Also, I highly recommend Paul Nahin's Inside Interesting Integrals. This book is incredibly well-written and a pleasure to read. It isn't exactly what you asked for, because it does eventually go above and beyond high school math. However: the first few chapters don't require much more than high school math. And in the later chapters, Nahin introduces the necessary concepts.



        Other sources:




        • Integral Kokeboken (http://folk.ntnu.no/oistes/Diverse/Integral%20Kokeboken.pdf) It is written in a language I don't understand, but there are lots of fun practice problems.


        • Advanced Integration Techniques (http://advancedintegrals.com/advanced-integration-techniques.pdf) I didn't like this as much as Nahin's book, but it covers some of the same things, and maybe it's more your thing.





        share|cite|improve this answer














        If you're looking for more practice problems, look for problems from MIT Integration Bees. Some of their problems are kind of easy, but there are quite a few gems.



        Also, I highly recommend Paul Nahin's Inside Interesting Integrals. This book is incredibly well-written and a pleasure to read. It isn't exactly what you asked for, because it does eventually go above and beyond high school math. However: the first few chapters don't require much more than high school math. And in the later chapters, Nahin introduces the necessary concepts.



        Other sources:




        • Integral Kokeboken (http://folk.ntnu.no/oistes/Diverse/Integral%20Kokeboken.pdf) It is written in a language I don't understand, but there are lots of fun practice problems.


        • Advanced Integration Techniques (http://advancedintegrals.com/advanced-integration-techniques.pdf) I didn't like this as much as Nahin's book, but it covers some of the same things, and maybe it's more your thing.






        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 3 at 15:14

























        answered Sep 3 at 15:11









        aras

        4,066732




        4,066732











        • any book that you would recommend for practising integration ?
          – JIM
          Sep 3 at 15:12










        • @JIM Nahin's book has some good practice problems. For a lot of problems. I'm pretty sure Integral Kokeboken has some more elementary stuff.
          – aras
          Sep 3 at 15:15










        • Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
          – JIM
          Sep 3 at 15:21










        • @JIM It's an actual book, so I don't have a link. You can buy it off of amazon (amazon.com/…) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
          – aras
          Sep 3 at 15:25






        • 4




          @JIM No, I'm not going to give you a link to the PDF since I'm not sure about StackExchange's policies. If you want a PDF, google resources for getting pdfs of textbooks. There are piracy websites that are easy enough to find and use.
          – aras
          Sep 3 at 15:29
















        • any book that you would recommend for practising integration ?
          – JIM
          Sep 3 at 15:12










        • @JIM Nahin's book has some good practice problems. For a lot of problems. I'm pretty sure Integral Kokeboken has some more elementary stuff.
          – aras
          Sep 3 at 15:15










        • Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
          – JIM
          Sep 3 at 15:21










        • @JIM It's an actual book, so I don't have a link. You can buy it off of amazon (amazon.com/…) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
          – aras
          Sep 3 at 15:25






        • 4




          @JIM No, I'm not going to give you a link to the PDF since I'm not sure about StackExchange's policies. If you want a PDF, google resources for getting pdfs of textbooks. There are piracy websites that are easy enough to find and use.
          – aras
          Sep 3 at 15:29















        any book that you would recommend for practising integration ?
        – JIM
        Sep 3 at 15:12




        any book that you would recommend for practising integration ?
        – JIM
        Sep 3 at 15:12












        @JIM Nahin's book has some good practice problems. For a lot of problems. I'm pretty sure Integral Kokeboken has some more elementary stuff.
        – aras
        Sep 3 at 15:15




        @JIM Nahin's book has some good practice problems. For a lot of problems. I'm pretty sure Integral Kokeboken has some more elementary stuff.
        – aras
        Sep 3 at 15:15












        Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
        – JIM
        Sep 3 at 15:21




        Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
        – JIM
        Sep 3 at 15:21












        @JIM It's an actual book, so I don't have a link. You can buy it off of amazon (amazon.com/…) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
        – aras
        Sep 3 at 15:25




        @JIM It's an actual book, so I don't have a link. You can buy it off of amazon (amazon.com/…) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
        – aras
        Sep 3 at 15:25




        4




        4




        @JIM No, I'm not going to give you a link to the PDF since I'm not sure about StackExchange's policies. If you want a PDF, google resources for getting pdfs of textbooks. There are piracy websites that are easy enough to find and use.
        – aras
        Sep 3 at 15:29




        @JIM No, I'm not going to give you a link to the PDF since I'm not sure about StackExchange's policies. If you want a PDF, google resources for getting pdfs of textbooks. There are piracy websites that are easy enough to find and use.
        – aras
        Sep 3 at 15:29










        up vote
        5
        down vote













        I see that nobody has mentioned a Euler Substitution. So I’ll go ahead and add that here. An Euler Substitution is essentially some kind of algebraic substitution for evaluating integrals of the form$$intmathrm dx, Rleft(x,sqrtax^2+bx+cright)$$



        First Substitution of Euler: If $a>0$ then make the substitution$$sqrtax^2+bx+c=pm xsqrt a+t$$where either the positive and negative sign can be chosen.



        Second Substitution of Euler: If $c>0$ then let$$sqrtax^2+bx+c=xtpmsqrt c$$Solve for $x$ and differentiate to find what $mathrm dx$ is equal to.$$x=frac pm2tsqrt c-ba-t^2$$



        Third Substitution of Euler: If the polynomial inside the square root has real roots $alpha$ and $beta$, then let$$sqrtax^2+bx+c=sqrta(x-alpha)(x-beta)=(x-alpha)t$$Therefore$$x=frac alphabeta-alpha t^2a-t^2$$



        Now let’s take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, it’s evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but I’ll use the first substitution because it’s nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$






        share|cite|improve this answer




















        • +1 ,I remember seeing these somewhere, thanks for reminding me!
          – Alvin Lepik
          Sep 3 at 17:06














        up vote
        5
        down vote













        I see that nobody has mentioned a Euler Substitution. So I’ll go ahead and add that here. An Euler Substitution is essentially some kind of algebraic substitution for evaluating integrals of the form$$intmathrm dx, Rleft(x,sqrtax^2+bx+cright)$$



        First Substitution of Euler: If $a>0$ then make the substitution$$sqrtax^2+bx+c=pm xsqrt a+t$$where either the positive and negative sign can be chosen.



        Second Substitution of Euler: If $c>0$ then let$$sqrtax^2+bx+c=xtpmsqrt c$$Solve for $x$ and differentiate to find what $mathrm dx$ is equal to.$$x=frac pm2tsqrt c-ba-t^2$$



        Third Substitution of Euler: If the polynomial inside the square root has real roots $alpha$ and $beta$, then let$$sqrtax^2+bx+c=sqrta(x-alpha)(x-beta)=(x-alpha)t$$Therefore$$x=frac alphabeta-alpha t^2a-t^2$$



        Now let’s take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, it’s evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but I’ll use the first substitution because it’s nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$






        share|cite|improve this answer




















        • +1 ,I remember seeing these somewhere, thanks for reminding me!
          – Alvin Lepik
          Sep 3 at 17:06












        up vote
        5
        down vote










        up vote
        5
        down vote









        I see that nobody has mentioned a Euler Substitution. So I’ll go ahead and add that here. An Euler Substitution is essentially some kind of algebraic substitution for evaluating integrals of the form$$intmathrm dx, Rleft(x,sqrtax^2+bx+cright)$$



        First Substitution of Euler: If $a>0$ then make the substitution$$sqrtax^2+bx+c=pm xsqrt a+t$$where either the positive and negative sign can be chosen.



        Second Substitution of Euler: If $c>0$ then let$$sqrtax^2+bx+c=xtpmsqrt c$$Solve for $x$ and differentiate to find what $mathrm dx$ is equal to.$$x=frac pm2tsqrt c-ba-t^2$$



        Third Substitution of Euler: If the polynomial inside the square root has real roots $alpha$ and $beta$, then let$$sqrtax^2+bx+c=sqrta(x-alpha)(x-beta)=(x-alpha)t$$Therefore$$x=frac alphabeta-alpha t^2a-t^2$$



        Now let’s take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, it’s evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but I’ll use the first substitution because it’s nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$






        share|cite|improve this answer












        I see that nobody has mentioned a Euler Substitution. So I’ll go ahead and add that here. An Euler Substitution is essentially some kind of algebraic substitution for evaluating integrals of the form$$intmathrm dx, Rleft(x,sqrtax^2+bx+cright)$$



        First Substitution of Euler: If $a>0$ then make the substitution$$sqrtax^2+bx+c=pm xsqrt a+t$$where either the positive and negative sign can be chosen.



        Second Substitution of Euler: If $c>0$ then let$$sqrtax^2+bx+c=xtpmsqrt c$$Solve for $x$ and differentiate to find what $mathrm dx$ is equal to.$$x=frac pm2tsqrt c-ba-t^2$$



        Third Substitution of Euler: If the polynomial inside the square root has real roots $alpha$ and $beta$, then let$$sqrtax^2+bx+c=sqrta(x-alpha)(x-beta)=(x-alpha)t$$Therefore$$x=frac alphabeta-alpha t^2a-t^2$$



        Now let’s take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, it’s evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but I’ll use the first substitution because it’s nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 3 at 16:56









        Frank W.

        2,2311315




        2,2311315











        • +1 ,I remember seeing these somewhere, thanks for reminding me!
          – Alvin Lepik
          Sep 3 at 17:06
















        • +1 ,I remember seeing these somewhere, thanks for reminding me!
          – Alvin Lepik
          Sep 3 at 17:06















        +1 ,I remember seeing these somewhere, thanks for reminding me!
        – Alvin Lepik
        Sep 3 at 17:06




        +1 ,I remember seeing these somewhere, thanks for reminding me!
        – Alvin Lepik
        Sep 3 at 17:06










        up vote
        4
        down vote













        Another neat trick is to add different forms of integrals to obtain a much simpler one.



        For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=int_-1^1frac11+f(x),dx$$ then we could replace $x$ by $-x$ giving $$I=-int_1^-1frac11+f(-x),dx=int_-1^1fracf(x)1+f(x),dx$$ and adding gives $$2I=int_-1^1,dx.$$






        share|cite|improve this answer
























          up vote
          4
          down vote













          Another neat trick is to add different forms of integrals to obtain a much simpler one.



          For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=int_-1^1frac11+f(x),dx$$ then we could replace $x$ by $-x$ giving $$I=-int_1^-1frac11+f(-x),dx=int_-1^1fracf(x)1+f(x),dx$$ and adding gives $$2I=int_-1^1,dx.$$






          share|cite|improve this answer






















            up vote
            4
            down vote










            up vote
            4
            down vote









            Another neat trick is to add different forms of integrals to obtain a much simpler one.



            For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=int_-1^1frac11+f(x),dx$$ then we could replace $x$ by $-x$ giving $$I=-int_1^-1frac11+f(-x),dx=int_-1^1fracf(x)1+f(x),dx$$ and adding gives $$2I=int_-1^1,dx.$$






            share|cite|improve this answer












            Another neat trick is to add different forms of integrals to obtain a much simpler one.



            For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=int_-1^1frac11+f(x),dx$$ then we could replace $x$ by $-x$ giving $$I=-int_1^-1frac11+f(-x),dx=int_-1^1fracf(x)1+f(x),dx$$ and adding gives $$2I=int_-1^1,dx.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 3 at 18:50









            TheSimpliFire

            11k62255




            11k62255




















                up vote
                3
                down vote













                One very simple example that pops to mind (several of my students have been dumbfounded on first sight with the following example).



                Find the anti derivative (there are many such examples)
                $$int fractt+1dt $$
                You add and subtract $1$ in the numerator and voila.




                Another one. Bearing in mind $int frac11+t^2dt = arctan t + C$. Find the anti derivative, say,
                $$int frac19t^2 -6t +2dt $$
                Separate the square part and integrate with respect to $3t-1$. Simplifies to finding
                $$frac13int frac1(3t-1)^2 +1d(3t-1) $$






                share|cite|improve this answer






















                • some other examples could you quote please ?
                  – JIM
                  Sep 3 at 15:13










                • @JIM I would have posted a similar example, but you mentioned that you are familiar with "partial fractions", which this kinda boils down too, so I thought this would be nothing new.
                  – Cornman
                  Sep 3 at 15:14










                • @Cornman in my experience, the less experienced practitioner starts trying to solve the problem using unnecessary force. Something along the lines of hunting down a fly with a cannon
                  – Alvin Lepik
                  Sep 3 at 15:20










                • @AlvinLepik You are absolutely correct .
                  – JIM
                  Sep 3 at 15:24














                up vote
                3
                down vote













                One very simple example that pops to mind (several of my students have been dumbfounded on first sight with the following example).



                Find the anti derivative (there are many such examples)
                $$int fractt+1dt $$
                You add and subtract $1$ in the numerator and voila.




                Another one. Bearing in mind $int frac11+t^2dt = arctan t + C$. Find the anti derivative, say,
                $$int frac19t^2 -6t +2dt $$
                Separate the square part and integrate with respect to $3t-1$. Simplifies to finding
                $$frac13int frac1(3t-1)^2 +1d(3t-1) $$






                share|cite|improve this answer






















                • some other examples could you quote please ?
                  – JIM
                  Sep 3 at 15:13










                • @JIM I would have posted a similar example, but you mentioned that you are familiar with "partial fractions", which this kinda boils down too, so I thought this would be nothing new.
                  – Cornman
                  Sep 3 at 15:14










                • @Cornman in my experience, the less experienced practitioner starts trying to solve the problem using unnecessary force. Something along the lines of hunting down a fly with a cannon
                  – Alvin Lepik
                  Sep 3 at 15:20










                • @AlvinLepik You are absolutely correct .
                  – JIM
                  Sep 3 at 15:24












                up vote
                3
                down vote










                up vote
                3
                down vote









                One very simple example that pops to mind (several of my students have been dumbfounded on first sight with the following example).



                Find the anti derivative (there are many such examples)
                $$int fractt+1dt $$
                You add and subtract $1$ in the numerator and voila.




                Another one. Bearing in mind $int frac11+t^2dt = arctan t + C$. Find the anti derivative, say,
                $$int frac19t^2 -6t +2dt $$
                Separate the square part and integrate with respect to $3t-1$. Simplifies to finding
                $$frac13int frac1(3t-1)^2 +1d(3t-1) $$






                share|cite|improve this answer














                One very simple example that pops to mind (several of my students have been dumbfounded on first sight with the following example).



                Find the anti derivative (there are many such examples)
                $$int fractt+1dt $$
                You add and subtract $1$ in the numerator and voila.




                Another one. Bearing in mind $int frac11+t^2dt = arctan t + C$. Find the anti derivative, say,
                $$int frac19t^2 -6t +2dt $$
                Separate the square part and integrate with respect to $3t-1$. Simplifies to finding
                $$frac13int frac1(3t-1)^2 +1d(3t-1) $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Sep 3 at 15:15

























                answered Sep 3 at 15:05









                Alvin Lepik

                2,561921




                2,561921











                • some other examples could you quote please ?
                  – JIM
                  Sep 3 at 15:13










                • @JIM I would have posted a similar example, but you mentioned that you are familiar with "partial fractions", which this kinda boils down too, so I thought this would be nothing new.
                  – Cornman
                  Sep 3 at 15:14










                • @Cornman in my experience, the less experienced practitioner starts trying to solve the problem using unnecessary force. Something along the lines of hunting down a fly with a cannon
                  – Alvin Lepik
                  Sep 3 at 15:20










                • @AlvinLepik You are absolutely correct .
                  – JIM
                  Sep 3 at 15:24
















                • some other examples could you quote please ?
                  – JIM
                  Sep 3 at 15:13










                • @JIM I would have posted a similar example, but you mentioned that you are familiar with "partial fractions", which this kinda boils down too, so I thought this would be nothing new.
                  – Cornman
                  Sep 3 at 15:14










                • @Cornman in my experience, the less experienced practitioner starts trying to solve the problem using unnecessary force. Something along the lines of hunting down a fly with a cannon
                  – Alvin Lepik
                  Sep 3 at 15:20










                • @AlvinLepik You are absolutely correct .
                  – JIM
                  Sep 3 at 15:24















                some other examples could you quote please ?
                – JIM
                Sep 3 at 15:13




                some other examples could you quote please ?
                – JIM
                Sep 3 at 15:13












                @JIM I would have posted a similar example, but you mentioned that you are familiar with "partial fractions", which this kinda boils down too, so I thought this would be nothing new.
                – Cornman
                Sep 3 at 15:14




                @JIM I would have posted a similar example, but you mentioned that you are familiar with "partial fractions", which this kinda boils down too, so I thought this would be nothing new.
                – Cornman
                Sep 3 at 15:14












                @Cornman in my experience, the less experienced practitioner starts trying to solve the problem using unnecessary force. Something along the lines of hunting down a fly with a cannon
                – Alvin Lepik
                Sep 3 at 15:20




                @Cornman in my experience, the less experienced practitioner starts trying to solve the problem using unnecessary force. Something along the lines of hunting down a fly with a cannon
                – Alvin Lepik
                Sep 3 at 15:20












                @AlvinLepik You are absolutely correct .
                – JIM
                Sep 3 at 15:24




                @AlvinLepik You are absolutely correct .
                – JIM
                Sep 3 at 15:24










                up vote
                2
                down vote













                Maybe something like this:



                Calculate $intln(x), dx$ where the trick is to write $int 1cdot ln(x), dx$ and then use partial integration.



                And other one, which you probably already know are integrals of this form:



                $int fracf'(x)f(x), dx=ln(|f(x)|)+c$



                Which finds use in many other integrals, often combined with other methods.






                share|cite|improve this answer
























                  up vote
                  2
                  down vote













                  Maybe something like this:



                  Calculate $intln(x), dx$ where the trick is to write $int 1cdot ln(x), dx$ and then use partial integration.



                  And other one, which you probably already know are integrals of this form:



                  $int fracf'(x)f(x), dx=ln(|f(x)|)+c$



                  Which finds use in many other integrals, often combined with other methods.






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    Maybe something like this:



                    Calculate $intln(x), dx$ where the trick is to write $int 1cdot ln(x), dx$ and then use partial integration.



                    And other one, which you probably already know are integrals of this form:



                    $int fracf'(x)f(x), dx=ln(|f(x)|)+c$



                    Which finds use in many other integrals, often combined with other methods.






                    share|cite|improve this answer












                    Maybe something like this:



                    Calculate $intln(x), dx$ where the trick is to write $int 1cdot ln(x), dx$ and then use partial integration.



                    And other one, which you probably already know are integrals of this form:



                    $int fracf'(x)f(x), dx=ln(|f(x)|)+c$



                    Which finds use in many other integrals, often combined with other methods.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 3 at 15:17









                    Cornman

                    2,84421228




                    2,84421228




















                        up vote
                        2
                        down vote













                        Another different trick suppose you want to calculate



                        $$I=int_-infty^inftye^-x²dx$$



                        It seems hard integration until you square it and use polar coordinates.
                        $x=rsin theta$ and $y=rcos theta$
                        $$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          Another different trick suppose you want to calculate



                          $$I=int_-infty^inftye^-x²dx$$



                          It seems hard integration until you square it and use polar coordinates.
                          $x=rsin theta$ and $y=rcos theta$
                          $$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            Another different trick suppose you want to calculate



                            $$I=int_-infty^inftye^-x²dx$$



                            It seems hard integration until you square it and use polar coordinates.
                            $x=rsin theta$ and $y=rcos theta$
                            $$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$






                            share|cite|improve this answer












                            Another different trick suppose you want to calculate



                            $$I=int_-infty^inftye^-x²dx$$



                            It seems hard integration until you square it and use polar coordinates.
                            $x=rsin theta$ and $y=rcos theta$
                            $$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 3 at 15:29









                            Deepesh Meena

                            4,43821025




                            4,43821025




















                                up vote
                                2
                                down vote













                                If you integrate even function $f$ on interval which is symmetric around zero, then you could just integrate $f$ on positive (or negative) part of interval. In other words, if $f(-x)=f(x)$ holds, then



                                $$intlimits_-a^af(x);mathrmdx = 2intlimits_0^af(x);mathrmdx$$



                                Analogously, integrating odd function $g$ on interval which is symmetric around zero will give you $0$:



                                $$intlimits_-a^ag(x);mathrmdx = 0$$



                                If function $h$ is periodic, with period $T$, then



                                $$intlimits_a^bh(x);mathrmdx = intlimits_a+T^b+Th(x);mathrmdx$$
                                and
                                $$intlimits_a^a+Th(x);mathrmdx = intlimits_b^b+Th(x);mathrmdx$$
                                for every real $a$ and $b$.



                                You could easily prove any of these statements.






                                share|cite|improve this answer




















                                • 1 and 2nd result i know but not able to get 3 and 4 one
                                  – JIM
                                  Sep 3 at 16:23










                                • Use integration by substitution.
                                  – Nikola Ubavić
                                  Sep 3 at 16:27










                                • no , I am not asking for the proof but what exactly does this result say ?
                                  – JIM
                                  Sep 3 at 16:31










                                • @JIM he's saying it's a function which repeats, like $ h(x) = sin(x) $. That has period $ T = pi $, so you can integrate over any different repeated period and get the same result or over any interval of length equal to the period.
                                  – theREALyumdub
                                  Sep 3 at 19:28















                                up vote
                                2
                                down vote













                                If you integrate even function $f$ on interval which is symmetric around zero, then you could just integrate $f$ on positive (or negative) part of interval. In other words, if $f(-x)=f(x)$ holds, then



                                $$intlimits_-a^af(x);mathrmdx = 2intlimits_0^af(x);mathrmdx$$



                                Analogously, integrating odd function $g$ on interval which is symmetric around zero will give you $0$:



                                $$intlimits_-a^ag(x);mathrmdx = 0$$



                                If function $h$ is periodic, with period $T$, then



                                $$intlimits_a^bh(x);mathrmdx = intlimits_a+T^b+Th(x);mathrmdx$$
                                and
                                $$intlimits_a^a+Th(x);mathrmdx = intlimits_b^b+Th(x);mathrmdx$$
                                for every real $a$ and $b$.



                                You could easily prove any of these statements.






                                share|cite|improve this answer




















                                • 1 and 2nd result i know but not able to get 3 and 4 one
                                  – JIM
                                  Sep 3 at 16:23










                                • Use integration by substitution.
                                  – Nikola Ubavić
                                  Sep 3 at 16:27










                                • no , I am not asking for the proof but what exactly does this result say ?
                                  – JIM
                                  Sep 3 at 16:31










                                • @JIM he's saying it's a function which repeats, like $ h(x) = sin(x) $. That has period $ T = pi $, so you can integrate over any different repeated period and get the same result or over any interval of length equal to the period.
                                  – theREALyumdub
                                  Sep 3 at 19:28













                                up vote
                                2
                                down vote










                                up vote
                                2
                                down vote









                                If you integrate even function $f$ on interval which is symmetric around zero, then you could just integrate $f$ on positive (or negative) part of interval. In other words, if $f(-x)=f(x)$ holds, then



                                $$intlimits_-a^af(x);mathrmdx = 2intlimits_0^af(x);mathrmdx$$



                                Analogously, integrating odd function $g$ on interval which is symmetric around zero will give you $0$:



                                $$intlimits_-a^ag(x);mathrmdx = 0$$



                                If function $h$ is periodic, with period $T$, then



                                $$intlimits_a^bh(x);mathrmdx = intlimits_a+T^b+Th(x);mathrmdx$$
                                and
                                $$intlimits_a^a+Th(x);mathrmdx = intlimits_b^b+Th(x);mathrmdx$$
                                for every real $a$ and $b$.



                                You could easily prove any of these statements.






                                share|cite|improve this answer












                                If you integrate even function $f$ on interval which is symmetric around zero, then you could just integrate $f$ on positive (or negative) part of interval. In other words, if $f(-x)=f(x)$ holds, then



                                $$intlimits_-a^af(x);mathrmdx = 2intlimits_0^af(x);mathrmdx$$



                                Analogously, integrating odd function $g$ on interval which is symmetric around zero will give you $0$:



                                $$intlimits_-a^ag(x);mathrmdx = 0$$



                                If function $h$ is periodic, with period $T$, then



                                $$intlimits_a^bh(x);mathrmdx = intlimits_a+T^b+Th(x);mathrmdx$$
                                and
                                $$intlimits_a^a+Th(x);mathrmdx = intlimits_b^b+Th(x);mathrmdx$$
                                for every real $a$ and $b$.



                                You could easily prove any of these statements.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Sep 3 at 16:12









                                Nikola Ubavić

                                805




                                805











                                • 1 and 2nd result i know but not able to get 3 and 4 one
                                  – JIM
                                  Sep 3 at 16:23










                                • Use integration by substitution.
                                  – Nikola Ubavić
                                  Sep 3 at 16:27










                                • no , I am not asking for the proof but what exactly does this result say ?
                                  – JIM
                                  Sep 3 at 16:31










                                • @JIM he's saying it's a function which repeats, like $ h(x) = sin(x) $. That has period $ T = pi $, so you can integrate over any different repeated period and get the same result or over any interval of length equal to the period.
                                  – theREALyumdub
                                  Sep 3 at 19:28

















                                • 1 and 2nd result i know but not able to get 3 and 4 one
                                  – JIM
                                  Sep 3 at 16:23










                                • Use integration by substitution.
                                  – Nikola Ubavić
                                  Sep 3 at 16:27










                                • no , I am not asking for the proof but what exactly does this result say ?
                                  – JIM
                                  Sep 3 at 16:31










                                • @JIM he's saying it's a function which repeats, like $ h(x) = sin(x) $. That has period $ T = pi $, so you can integrate over any different repeated period and get the same result or over any interval of length equal to the period.
                                  – theREALyumdub
                                  Sep 3 at 19:28
















                                1 and 2nd result i know but not able to get 3 and 4 one
                                – JIM
                                Sep 3 at 16:23




                                1 and 2nd result i know but not able to get 3 and 4 one
                                – JIM
                                Sep 3 at 16:23












                                Use integration by substitution.
                                – Nikola Ubavić
                                Sep 3 at 16:27




                                Use integration by substitution.
                                – Nikola Ubavić
                                Sep 3 at 16:27












                                no , I am not asking for the proof but what exactly does this result say ?
                                – JIM
                                Sep 3 at 16:31




                                no , I am not asking for the proof but what exactly does this result say ?
                                – JIM
                                Sep 3 at 16:31












                                @JIM he's saying it's a function which repeats, like $ h(x) = sin(x) $. That has period $ T = pi $, so you can integrate over any different repeated period and get the same result or over any interval of length equal to the period.
                                – theREALyumdub
                                Sep 3 at 19:28





                                @JIM he's saying it's a function which repeats, like $ h(x) = sin(x) $. That has period $ T = pi $, so you can integrate over any different repeated period and get the same result or over any interval of length equal to the period.
                                – theREALyumdub
                                Sep 3 at 19:28



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