Smart Integration Tricks [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
9
down vote
favorite
I am in the last year of my school
and studied integration this year
I have done several Integration techniques like Integration
- By substitution
- By partial fractions
- By parts
- Trigo. substitutions
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
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.
add a comment |Â
up vote
9
down vote
favorite
I am in the last year of my school
and studied integration this year
I have done several Integration techniques like Integration
- By substitution
- By partial fractions
- By parts
- Trigo. substitutions
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
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
â Michael Hoppe
Sep 3 at 18:35
For integration by parts have a look at math.stackexchange.com/questions/20397/â¦
â Michael Hoppe
Sep 3 at 18:39
Also related: math.stackexchange.com/questions/70974/â¦
â Hans Lundmark
Sep 3 at 20:22
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I am in the last year of my school
and studied integration this year
I have done several Integration techniques like Integration
- By substitution
- By partial fractions
- By parts
- Trigo. substitutions
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
I am in the last year of my school
and studied integration this year
I have done several Integration techniques like Integration
- By substitution
- By partial fractions
- By parts
- Trigo. substitutions
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
integration definite-integrals indefinite-integrals
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
â Michael Hoppe
Sep 3 at 18:35
For integration by parts have a look at math.stackexchange.com/questions/20397/â¦
â Michael Hoppe
Sep 3 at 18:39
Also related: math.stackexchange.com/questions/70974/â¦
â Hans Lundmark
Sep 3 at 20:22
add a comment |Â
Have a look here math.stackexchange.com/questions/2821112/integral-milking
â Michael Hoppe
Sep 3 at 18:35
For integration by parts have a look at math.stackexchange.com/questions/20397/â¦
â Michael Hoppe
Sep 3 at 18:39
Also related: math.stackexchange.com/questions/70974/â¦
â Hans Lundmark
Sep 3 at 20:22
Have a look here math.stackexchange.com/questions/2821112/integral-milking
â Michael Hoppe
Sep 3 at 18:35
Have a look here math.stackexchange.com/questions/2821112/integral-milking
â Michael Hoppe
Sep 3 at 18:35
For integration by parts have a look at math.stackexchange.com/questions/20397/â¦
â Michael Hoppe
Sep 3 at 18:39
For integration by parts have a look at math.stackexchange.com/questions/20397/â¦
â Michael Hoppe
Sep 3 at 18:39
Also related: math.stackexchange.com/questions/70974/â¦
â Hans Lundmark
Sep 3 at 20:22
Also related: math.stackexchange.com/questions/70974/â¦
â Hans Lundmark
Sep 3 at 20:22
add a comment |Â
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.
any book that you would recommend for practising integration ?
â 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.
â aras
Sep 3 at 15:15
Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
â 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/â¦) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
â 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.
â aras
Sep 3 at 15:29
 |Â
show 1 more comment
up vote
5
down vote
I see that nobody has mentioned a Euler Substitution. So IâÂÂll 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 letâÂÂs take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, itâÂÂs evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but IâÂÂll use the first substitution because itâÂÂs nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$
+1 ,I remember seeing these somewhere, thanks for reminding me!
â Alvin Lepik
Sep 3 at 17:06
add a comment |Â
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 $$2I=int_-1^1,dx.$$
add a comment |Â
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
$$frac13int frac1(3t-1)^2 +1d(3t-1) $$
some other examples could you quote please ?
â 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.
â 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
â Alvin Lepik
Sep 3 at 15:20
@AlvinLepik You are absolutely correct .
â JIM
Sep 3 at 15:24
add a comment |Â
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.
add a comment |Â
up vote
2
down vote
Another different trick suppose you want to calculate
$$I=int_-infty^inftye^-xòdx$$
It seems hard integration until you square it and use polar coordinates.
$x=rsin theta$ and $y=rcos theta$
$$left(int_-infty^inftye^-xòdxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$
add a comment |Â
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.
1 and 2nd result i know but not able to get 3 and 4 one
â JIM
Sep 3 at 16:23
Use integration by substitution.
â Nikola UbaviÃÂ
Sep 3 at 16:27
no , I am not asking for the proof but what exactly does this result say ?
â 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.
â theREALyumdub
Sep 3 at 19:28
add a comment |Â
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.
any book that you would recommend for practising integration ?
â 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.
â aras
Sep 3 at 15:15
Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
â 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/â¦) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
â 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.
â aras
Sep 3 at 15:29
 |Â
show 1 more comment
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.
any book that you would recommend for practising integration ?
â 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.
â aras
Sep 3 at 15:15
Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
â 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/â¦) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
â 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.
â aras
Sep 3 at 15:29
 |Â
show 1 more comment
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.
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.
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 ?
â 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.
â aras
Sep 3 at 15:15
Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
â 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/â¦) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
â 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.
â aras
Sep 3 at 15:29
 |Â
show 1 more comment
any book that you would recommend for practising integration ?
â 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.
â aras
Sep 3 at 15:15
Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
â 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/â¦) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
â 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.
â aras
Sep 3 at 15:29
any book that you would recommend for practising integration ?
â JIM
Sep 3 at 15:12
any book that you would recommend for practising integration ?
â 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.
â 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.
â aras
Sep 3 at 15:15
Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
â JIM
Sep 3 at 15:21
Could you please provide the link for Paul Nahin's Inside Interesting Integrals ?
â 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/â¦) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
â 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/â¦) or download the pdf from a piracy site. I did both since I wanted to support the author and like having a PDF.
â 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.
â 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.
â aras
Sep 3 at 15:29
 |Â
show 1 more comment
up vote
5
down vote
I see that nobody has mentioned a Euler Substitution. So IâÂÂll 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 letâÂÂs take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, itâÂÂs evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but IâÂÂll use the first substitution because itâÂÂs nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$
+1 ,I remember seeing these somewhere, thanks for reminding me!
â Alvin Lepik
Sep 3 at 17:06
add a comment |Â
up vote
5
down vote
I see that nobody has mentioned a Euler Substitution. So IâÂÂll 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 letâÂÂs take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, itâÂÂs evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but IâÂÂll use the first substitution because itâÂÂs nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$
+1 ,I remember seeing these somewhere, thanks for reminding me!
â Alvin Lepik
Sep 3 at 17:06
add a comment |Â
up vote
5
down vote
up vote
5
down vote
I see that nobody has mentioned a Euler Substitution. So IâÂÂll 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 letâÂÂs take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, itâÂÂs evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but IâÂÂll use the first substitution because itâÂÂs nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$
I see that nobody has mentioned a Euler Substitution. So IâÂÂll 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 letâÂÂs take an example. Say we wanted to evaluate the integral$$intfrac mathrm dxsqrtx^2+1$$Here, itâÂÂs evident that $a=c=1$ and $b=0$. Of course, we can use the second substitution, but IâÂÂll use the first substitution because itâÂÂs nicer. Take$$sqrtx^2+1=-x+t$$So that$$x=frac t^2-12tqquadqquadmathrm dx=frac t^2+12t^2,mathrm dt$$Hence$$intfrac mathrm dxsqrtx^2+1=intfrac mathrm dtt=logleft(x+sqrtx^2+1right)+C$$
answered Sep 3 at 16:56
Frank W.
2,2311315
2,2311315
+1 ,I remember seeing these somewhere, thanks for reminding me!
â Alvin Lepik
Sep 3 at 17:06
add a comment |Â
+1 ,I remember seeing these somewhere, thanks for reminding me!
â Alvin Lepik
Sep 3 at 17:06
+1 ,I remember seeing these somewhere, thanks for reminding me!
â Alvin Lepik
Sep 3 at 17:06
+1 ,I remember seeing these somewhere, thanks for reminding me!
â Alvin Lepik
Sep 3 at 17:06
add a comment |Â
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 $$2I=int_-1^1,dx.$$
add a comment |Â
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 $$2I=int_-1^1,dx.$$
add a comment |Â
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 $$2I=int_-1^1,dx.$$
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 $$2I=int_-1^1,dx.$$
answered Sep 3 at 18:50
TheSimpliFire
11k62255
11k62255
add a comment |Â
add a comment |Â
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
$$frac13int frac1(3t-1)^2 +1d(3t-1) $$
some other examples could you quote please ?
â 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.
â 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
â Alvin Lepik
Sep 3 at 15:20
@AlvinLepik You are absolutely correct .
â JIM
Sep 3 at 15:24
add a comment |Â
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
$$frac13int frac1(3t-1)^2 +1d(3t-1) $$
some other examples could you quote please ?
â 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.
â 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
â Alvin Lepik
Sep 3 at 15:20
@AlvinLepik You are absolutely correct .
â JIM
Sep 3 at 15:24
add a comment |Â
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
$$frac13int frac1(3t-1)^2 +1d(3t-1) $$
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
$$frac13int frac1(3t-1)^2 +1d(3t-1) $$
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 ?
â 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.
â 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
â Alvin Lepik
Sep 3 at 15:20
@AlvinLepik You are absolutely correct .
â JIM
Sep 3 at 15:24
add a comment |Â
some other examples could you quote please ?
â 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.
â 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
â Alvin Lepik
Sep 3 at 15:20
@AlvinLepik You are absolutely correct .
â JIM
Sep 3 at 15:24
some other examples could you quote please ?
â JIM
Sep 3 at 15:13
some other examples could you quote please ?
â 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.
â 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.
â 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
â 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
â Alvin Lepik
Sep 3 at 15:20
@AlvinLepik You are absolutely correct .
â JIM
Sep 3 at 15:24
@AlvinLepik You are absolutely correct .
â JIM
Sep 3 at 15:24
add a comment |Â
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.
add a comment |Â
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.
add a comment |Â
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.
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.
answered Sep 3 at 15:17
Cornman
2,84421228
2,84421228
add a comment |Â
add a comment |Â
up vote
2
down vote
Another different trick suppose you want to calculate
$$I=int_-infty^inftye^-xòdx$$
It seems hard integration until you square it and use polar coordinates.
$x=rsin theta$ and $y=rcos theta$
$$left(int_-infty^inftye^-xòdxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$
add a comment |Â
up vote
2
down vote
Another different trick suppose you want to calculate
$$I=int_-infty^inftye^-xòdx$$
It seems hard integration until you square it and use polar coordinates.
$x=rsin theta$ and $y=rcos theta$
$$left(int_-infty^inftye^-xòdxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Another different trick suppose you want to calculate
$$I=int_-infty^inftye^-xòdx$$
It seems hard integration until you square it and use polar coordinates.
$x=rsin theta$ and $y=rcos theta$
$$left(int_-infty^inftye^-xòdxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$
Another different trick suppose you want to calculate
$$I=int_-infty^inftye^-xòdx$$
It seems hard integration until you square it and use polar coordinates.
$x=rsin theta$ and $y=rcos theta$
$$left(int_-infty^inftye^-xòdxright )^2=int_mathbbR^2e^-(x^2+y^2)dx dy=int_0^2 piint_0^inftye^-r^2r dr dtheta=2piint_0^inftye^-r^2 rdr=pi$$
answered Sep 3 at 15:29
Deepesh Meena
4,43821025
4,43821025
add a comment |Â
add a comment |Â
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.
1 and 2nd result i know but not able to get 3 and 4 one
â JIM
Sep 3 at 16:23
Use integration by substitution.
â Nikola UbaviÃÂ
Sep 3 at 16:27
no , I am not asking for the proof but what exactly does this result say ?
â 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.
â theREALyumdub
Sep 3 at 19:28
add a comment |Â
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.
1 and 2nd result i know but not able to get 3 and 4 one
â JIM
Sep 3 at 16:23
Use integration by substitution.
â Nikola UbaviÃÂ
Sep 3 at 16:27
no , I am not asking for the proof but what exactly does this result say ?
â 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.
â theREALyumdub
Sep 3 at 19:28
add a comment |Â
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.
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.
answered Sep 3 at 16:12
Nikola UbaviÃÂ
805
805
1 and 2nd result i know but not able to get 3 and 4 one
â JIM
Sep 3 at 16:23
Use integration by substitution.
â Nikola UbaviÃÂ
Sep 3 at 16:27
no , I am not asking for the proof but what exactly does this result say ?
â 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.
â theREALyumdub
Sep 3 at 19:28
add a comment |Â
1 and 2nd result i know but not able to get 3 and 4 one
â JIM
Sep 3 at 16:23
Use integration by substitution.
â Nikola UbaviÃÂ
Sep 3 at 16:27
no , I am not asking for the proof but what exactly does this result say ?
â 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.
â theREALyumdub
Sep 3 at 19:28
1 and 2nd result i know but not able to get 3 and 4 one
â JIM
Sep 3 at 16:23
1 and 2nd result i know but not able to get 3 and 4 one
â JIM
Sep 3 at 16:23
Use integration by substitution.
â Nikola UbaviÃÂ
Sep 3 at 16:27
Use integration by substitution.
â Nikola UbaviÃÂ
Sep 3 at 16:27
no , I am not asking for the proof but what exactly does this result say ?
â JIM
Sep 3 at 16:31
no , I am not asking for the proof but what exactly does this result say ?
â 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.
â 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.
â theREALyumdub
Sep 3 at 19:28
add a comment |Â
Have a look here math.stackexchange.com/questions/2821112/integral-milking
â Michael Hoppe
Sep 3 at 18:35
For integration by parts have a look at math.stackexchange.com/questions/20397/â¦
â Michael Hoppe
Sep 3 at 18:39
Also related: math.stackexchange.com/questions/70974/â¦
â Hans Lundmark
Sep 3 at 20:22