Trig Subsitution When There's No Square Root
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$$
Anyway, so far, I have that:
$$x = rtan theta$$
$$dx = rsec^2 theta$$
$$sqrt (r^2+x^2) = rsectheta$$
The triangle I based the above values on:
Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.
beginalign
&= Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta \
&= Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta \
&= fracAr int_a^b frac1sec^4thetadtheta \
&= fracAr int_a^b cos^4theta dtheta \
&= fracAr int_a^b (cos^2theta)^2 dtheta \
&= fracAr int_a^b left[ frac12 1+cos(2theta)) right]^2dtheta \
&= fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta \
&= fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta
endalign
And from there it gets really messed up and I end up with a weird semi-final answer of $$fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$$ which is wrong after I make substitutions.
I already know that the final answer is $dfracArleft(1-dfracasqrtr^2+a^2right)$, but I really want to understand this.
calculus integration improper-integrals trigonometric-integrals
edited Mar 11 at 5:31
David K
55.6k345121
asked Mar 10 at 21:09
CodingMeeCodingMee
756
$endgroup$
add a comment |
$begingroup$
I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$$
Anyway, so far, I have that:
$$x = rtan theta$$
$$dx = rsec^2 theta$$
$$sqrt (r^2+x^2) = rsectheta$$
The triangle I based the above values on:
Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.
beginalign
&= Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta \
&= Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta \
&= fracAr int_a^b frac1sec^4thetadtheta \
&= fracAr int_a^b cos^4theta dtheta \
&= fracAr int_a^b (cos^2theta)^2 dtheta \
&= fracAr int_a^b left[ frac12 1+cos(2theta)) right]^2dtheta \
&= fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta \
&= fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta
endalign
And from there it gets really messed up and I end up with a weird semi-final answer of $$fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$$ which is wrong after I make substitutions.
I already know that the final answer is $dfracArleft(1-dfracasqrtr^2+a^2right)$, but I really want to understand this.
calculus integration improper-integrals trigonometric-integrals
edited Mar 11 at 5:31
David K
55.6k345121
asked Mar 10 at 21:09
CodingMeeCodingMee
756
$endgroup$
2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
Mar 10 at 21:22
add a comment |
$begingroup$
I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$$
Anyway, so far, I have that:
$$x = rtan theta$$
$$dx = rsec^2 theta$$
$$sqrt (r^2+x^2) = rsectheta$$
The triangle I based the above values on:
Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.
beginalign
&= Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta \
&= Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta \
&= fracAr int_a^b frac1sec^4thetadtheta \
&= fracAr int_a^b cos^4theta dtheta \
&= fracAr int_a^b (cos^2theta)^2 dtheta \
&= fracAr int_a^b left[ frac12 1+cos(2theta)) right]^2dtheta \
&= fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta \
&= fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta
endalign
And from there it gets really messed up and I end up with a weird semi-final answer of $$fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$$ which is wrong after I make substitutions.
I already know that the final answer is $dfracArleft(1-dfracasqrtr^2+a^2right)$, but I really want to understand this.
calculus integration improper-integrals trigonometric-integrals
edited Mar 11 at 5:31
David K
55.6k345121
asked Mar 10 at 21:09
CodingMeeCodingMee
756
$endgroup$
I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$$
Anyway, so far, I have that:
$$x = rtan theta$$
$$dx = rsec^2 theta$$
$$sqrt (r^2+x^2) = rsectheta$$
The triangle I based the above values on:
Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.
beginalign
&= Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta \
&= Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta \
&= fracAr int_a^b frac1sec^4thetadtheta \
&= fracAr int_a^b cos^4theta dtheta \
&= fracAr int_a^b (cos^2theta)^2 dtheta \
&= fracAr int_a^b left[ frac12 1+cos(2theta)) right]^2dtheta \
&= fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta \
&= fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta
endalign
And from there it gets really messed up and I end up with a weird semi-final answer of $$fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$$ which is wrong after I make substitutions.
I already know that the final answer is $dfracArleft(1-dfracasqrtr^2+a^2right)$, but I really want to understand this.
calculus integration improper-integrals trigonometric-integrals
calculus integration improper-integrals trigonometric-integrals
edited Mar 11 at 5:31
David K
55.6k345121
asked Mar 10 at 21:09
CodingMeeCodingMee
756
edited Mar 11 at 5:31
David K
55.6k345121
asked Mar 10 at 21:09
CodingMeeCodingMee
756
edited Mar 11 at 5:31
David K
55.6k345121
edited Mar 11 at 5:31
David K
55.6k345121
edited Mar 11 at 5:31
David K
55.6k345121
55.6k345121
asked Mar 10 at 21:09
CodingMeeCodingMee
756
asked Mar 10 at 21:09
CodingMeeCodingMee
756
asked Mar 10 at 21:09
CodingMeeCodingMee
756
756
2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
Mar 10 at 21:22
add a comment |
2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
Mar 10 at 21:22
2
2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
Mar 10 at 21:22
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
Mar 10 at 21:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
$$
Now let's concentrate on the antiderivative
$$
intfrac1(1+u^2)^3/2,du=
intfrac1+u^2-u^2(1+u^2)^3/2,du=
intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
$$
Do the second term by parts
$$
int ufracu(1+u^2)^3/2,du=
-fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
$$
See what happens?
$$
intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
$$
which we can verify by direct differentiation.
Now
$$
left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
=1-fraca(r^2+a^2)^1/2
$$
and your integral is indeed
$$
fracArleft(1-fracasqrtr^2+a^2right)
$$
answered Mar 10 at 21:57
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
Firstly you made an error in the first line of working
$$(rsec(theta))^3=r^3sec^3(theta)$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.
answered Mar 10 at 21:32
Peter ForemanPeter Foreman
6,4461317
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
$$
Now let's concentrate on the antiderivative
$$
intfrac1(1+u^2)^3/2,du=
intfrac1+u^2-u^2(1+u^2)^3/2,du=
intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
$$
Do the second term by parts
$$
int ufracu(1+u^2)^3/2,du=
-fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
$$
See what happens?
$$
intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
$$
which we can verify by direct differentiation.
Now
$$
left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
=1-fraca(r^2+a^2)^1/2
$$
and your integral is indeed
$$
fracArleft(1-fracasqrtr^2+a^2right)
$$
answered Mar 10 at 21:57
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
$$
Now let's concentrate on the antiderivative
$$
intfrac1(1+u^2)^3/2,du=
intfrac1+u^2-u^2(1+u^2)^3/2,du=
intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
$$
Do the second term by parts
$$
int ufracu(1+u^2)^3/2,du=
-fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
$$
See what happens?
$$
intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
$$
which we can verify by direct differentiation.
Now
$$
left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
=1-fraca(r^2+a^2)^1/2
$$
and your integral is indeed
$$
fracArleft(1-fracasqrtr^2+a^2right)
$$
answered Mar 10 at 21:57
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
$$
Now let's concentrate on the antiderivative
$$
intfrac1(1+u^2)^3/2,du=
intfrac1+u^2-u^2(1+u^2)^3/2,du=
intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
$$
Do the second term by parts
$$
int ufracu(1+u^2)^3/2,du=
-fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
$$
See what happens?
$$
intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
$$
which we can verify by direct differentiation.
Now
$$
left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
=1-fraca(r^2+a^2)^1/2
$$
and your integral is indeed
$$
fracArleft(1-fracasqrtr^2+a^2right)
$$
answered Mar 10 at 21:57
egregegreg
185k1486208
$endgroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
$$
Now let's concentrate on the antiderivative
$$
intfrac1(1+u^2)^3/2,du=
intfrac1+u^2-u^2(1+u^2)^3/2,du=
intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
$$
Do the second term by parts
$$
int ufracu(1+u^2)^3/2,du=
-fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
$$
See what happens?
$$
intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
$$
which we can verify by direct differentiation.
Now
$$
left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
=1-fraca(r^2+a^2)^1/2
$$
and your integral is indeed
$$
fracArleft(1-fracasqrtr^2+a^2right)
$$
answered Mar 10 at 21:57
egregegreg
185k1486208
answered Mar 10 at 21:57
egregegreg
185k1486208
answered Mar 10 at 21:57
egregegreg
185k1486208
answered Mar 10 at 21:57
egregegreg
185k1486208
185k1486208
add a comment |
add a comment |
$begingroup$
Firstly you made an error in the first line of working
$$(rsec(theta))^3=r^3sec^3(theta)$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.
answered Mar 10 at 21:32
Peter ForemanPeter Foreman
6,4461317
$endgroup$
add a comment |
$begingroup$
Firstly you made an error in the first line of working
$$(rsec(theta))^3=r^3sec^3(theta)$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.
answered Mar 10 at 21:32
Peter ForemanPeter Foreman
6,4461317
$endgroup$
add a comment |
$begingroup$
Firstly you made an error in the first line of working
$$(rsec(theta))^3=r^3sec^3(theta)$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.
answered Mar 10 at 21:32
Peter ForemanPeter Foreman
6,4461317
$endgroup$
Firstly you made an error in the first line of working
$$(rsec(theta))^3=r^3sec^3(theta)$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.
answered Mar 10 at 21:32
Peter ForemanPeter Foreman
6,4461317
answered Mar 10 at 21:32
Peter ForemanPeter Foreman
6,4461317
answered Mar 10 at 21:32
Peter ForemanPeter Foreman
6,4461317
answered Mar 10 at 21:32
Peter ForemanPeter Foreman
6,4461317
6,4461317
add a comment |
add a comment |
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$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
Mar 10 at 21:22