# Trig sub integrals

• Sep 9th 2008, 12:21 PM
symstar
Trig sub integrals
$\displaystyle \int x\sqrt{1-x^{4}}dx$

Here's the work I've done, hopefully there's not too much bad math in it...
$\displaystyle x=\sin{\theta}$ so $\displaystyle dx=\cos{\theta}d{\theta}$

$\displaystyle \int \sin{\theta}\sqrt{(1-\sin^2{\theta})^2}\cos{\theta}d{\theta}$
$\displaystyle \int \sin{\theta}\sqrt{(\cos^2{\theta})^2}\cos{\theta}d {\theta}$
$\displaystyle \int \sin{\theta}\cos^2{\theta}\cos{\theta}d{\theta}$
$\displaystyle u=\sin{\theta}$ so $\displaystyle du=\cos{\theta}d{\theta}$
$\displaystyle \int u(1-u^2)du$
$\displaystyle \tfrac{1}{2}u^2 - \tfrac{1}{4}u^4 + C$

...and I stopped right here, seeing that this wouldn't give me the right answer. I'm not ever certain that I can u-sub in this problem like I did. Can someone point me in the right direction please?
• Sep 9th 2008, 12:42 PM
Shyam
$\displaystyle \int {x\sqrt {1 - x^4 } } dx \hfill \\$

$\displaystyle {\text{Let us substitute, }}x^2 = \sin \theta \Rightarrow \theta = \sin ^{ - 1} \left( {x^2 } \right) \hfill \\$

$\displaystyle {\text{so that, }}2xdx = \cos \theta d\theta \Rightarrow xdx = \frac{1} {2}\cos \theta d\theta \hfill \\$

$\displaystyle {\text{Now see, }} = \frac{1} {2}\int {\left( {\sqrt {1 - \sin ^2 \theta } } \right)} \cos \theta ~d\theta = \frac{1} {2}\int {\cos \theta \sqrt {\cos ^2 \theta } } ~d\theta \hfill \\$

$\displaystyle = \frac{1} {2}\int {\cos ^2 \theta ~d\theta } \hfill \\$

$\displaystyle = \frac{1} {2}\int {\frac{{1 + \cos 2\theta }} {2}} ~d\theta \hfill \\$

$\displaystyle = \frac{1} {4}\left( {\theta + \sin 2\theta } \right) + c \hfill \\$

$\displaystyle = \frac{1} {4}\left( {\theta + 2\sin \theta \cos \theta } \right) + c \hfill \\$

$\displaystyle = \frac{1} {4}\left( {\theta + 2\sin \theta \sqrt {1 - \sin ^2 \theta } } \right) + c \hfill \\$

$\displaystyle = \frac{1} {4}\left[ {\sin ^{ - 1} \left( {x^2 } \right) + 2x^2 \sqrt {1 - x^4 } } \right] + c \hfill \\$
• Sep 9th 2008, 01:56 PM
11rdc11
Quote:

Originally Posted by symstar
$\displaystyle \int x\sqrt{1-x^{4}}dx$

Here's the work I've done, hopefully there's not too much bad math in it...
$\displaystyle x=\sin{\theta}$ so $\displaystyle dx=\cos{\theta}d{\theta}$

$\displaystyle \int \sin{\theta}\sqrt{(1-\sin^2{\theta})^2}\cos{\theta}d{\theta}$

Using $\displaystyle x = sin\theta$

gives

$\displaystyle 1-sin^4\theta = (1 + sin^2\theta)(1-sin^2\theta)$

not

$\displaystyle (1-sin^2\theta)^2$

Hope that helps
• Sep 9th 2008, 04:42 PM
symstar
Quote:

Originally Posted by 11rdc11
Using $\displaystyle x = sin\theta$

gives

$\displaystyle 1-sin^4\theta = (1 + sin^2\theta)(1-sin^2\theta)$

not

$\displaystyle (1-sin^2\theta)^2$

Hope that helps

It does help and hopefully I won't do bad math like this in the future. (Yes)

Also, thank you Shyam for making sense of this problem for me. I did just want to point out that I'm pretty sure that
$\displaystyle = \frac{1} {2}\int {\frac{{1 + \cos 2\theta }} {2}} ~d\theta \hfill$
actually gives
$\displaystyle = \frac{1} {4}\left( {\theta + \tfrac{1}{2}\sin 2\theta } \right) + c \hfill$

but it doesn't change the answer much and it was still a lot of help.
• Sep 9th 2008, 06:09 PM
11rdc11
Quote:

Originally Posted by symstar
It does help and hopefully I won't do bad math like this in the future. (Yes)

Also, thank you Shyam for making sense of this problem for me. I did just want to point out that I'm pretty sure that
$\displaystyle = \frac{1} {2}\int {\frac{{1 + \cos 2\theta }} {2}} ~d\theta \hfill$
actually gives
$\displaystyle = \frac{1} {4}\left( {\theta + \tfrac{1}{2}\sin 2\theta } \right) + c \hfill$

but it doesn't change the answer much and it was still a lot of help.

Yep you are correct