# Math Help - Trig sub integrals

1. ## Trig sub integrals

$\int x\sqrt{1-x^{4}}dx$

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

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

...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?

2. $\int {x\sqrt {1 - x^4 } } dx \hfill \\$

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

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

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

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

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

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

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

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

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

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

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

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

gives

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

not

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

Hope that helps

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

gives

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

not

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

Hope that helps
It does help and hopefully I won't do bad math like this in the future.

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

actually gives
$= \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.

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

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

actually gives
$= \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