# Thread: How do I integrate this?

1. ## How do I integrate this?

I am stuck!

$\int x^{3}\cos{\tan^{-1}{x}}$

I tried

$\mbox{Let } u=\tan^{-1}{x}, du=\frac{1}{x^{2} +1}$

Then with some substituting,

$\int x^{3}\cos{\tan^{-1}{x}} = \int\sin^{3}{u}$

but I don't know how to integrate the second form any more than the first. Can anyone explain how to do this?

2. Originally Posted by sinewave85
I am stuck!

$\int x^{3}\cos{\tan^{-1}{x}}$

I tried

$\mbox{Let } u=\tan^{-1}{x}, du=\frac{1}{x^{2} +1}$

Then with some substituting,

$\int x^{3}\cos{\tan^{-1}{x}} = \int\sin^{3}{u}$

but I don't know how to integrate the second form any more than the first. Can anyone explain how to do this?

$\int \sin^3(u)du=\int (1-\cos^2(u))\sin(u)du$

Now let $w=\sin(u) \implies dw=-\sin(u)du$ and you are off to the races

3. Originally Posted by TheEmptySet
$\int \sin^3(u)du=\int (1-\cos^2(u))\sin(u)du$

Now let $w=\sin(u) \implies dw=-\sin(u)du$ and you are off to the races
$\mbox{But isn't }dw=d(sin(u))=cos(u)du?$

4. I scared my helper away

I still am a little confused.

If you let $w=1-\cos^{2}{u}\implies dw=2\cos{u}\sin{u}\implies u=\cos^{-1}{\sqrt{1-w}}$

Which makes it

$\int\frac{w}{2\sqrt{1-w}}dw$

Then

$z=1-w\implies dz=-1dw \implies w=1-2$

$\frac{-1}{2}\int\frac{1-z}{z^{1/2}}dz$

$\frac{-1}{2}\int z^{-1/2} - z^{1/2}dz$

And then integrate and back-substitute?

5. Hi sinewave85

I can't get : $\int x^{3}\cos{\tan^{-1}{x}} ~dx= \int\sin^{3}{u}~du$

Can you tell me how you got it?

6. Originally Posted by songoku
Hi sinewave85

I can't get : $\int x^{3}\cos{\tan^{-1}{x}} ~dx= \int\sin^{3}{u}~du$

Can you tell me how you got it?
Sure! (I hope this is legit)

$
\mbox{Let } u=\tan^{-1}{x}\implies du=\frac{1}{x^{2} +1}, u=\tan^{-1}{x}\implies \tan{u}=x$

thus

$
\int x^{3}\cos{\tan^{-1}{x}}~dx = \int\frac{x^{3}\cos{u}}{x^{2}+1}du
$

$
\int x^{3}\cos{\tan^{-1}{x}}~dx = \int\frac{tan^{3}{u}\cos{u}}{tan^{2}{u}+1}du
$

$
\int x^{3}\cos{\tan^{-1}{x}}~dx = \int\frac{tan^{3}{u}\cos{u}}{sec^{2}{u}}du
$

$
\int x^{3}\cos{\tan^{-1}{x}} ~dx= \int\sin^{3}{u}~du
$

There you go! I have to admit, this was a real challenge to work through!

7. Hi sinewave85
Originally Posted by sinewave85
$
\mbox{Let } u=\tan^{-1}{x}\implies du=\frac{1}{x^{2} +1}, u=\tan^{-1}{x}\implies \tan{u}=x$

thus

$
\int x^{3}\cos{\tan^{-1}{x}}~dx = \int\frac{x^{3}\cos{u}}{x^{2}+1}du
$

You made mistake when substituting dx. You got $du=\frac{1}{x^{2} +1}~dx$ , so $dx=(x^2+1)~du$