1. ## Calculating intregral

I've been trying to calculating the following integral
$\int xsin^{-1}(x)\, dx$

Here's my solution as far as I've been able to calculate. Let
$U=sin^{-1}x \Rightarrow dU=\frac{1}{\sqrt{1-x^2}}dx$
$dV = x \cdot dx \Rightarrow V = \frac{x^2}{2}$
Then
$I=VU-\int V\cdot du=sin^{-1}x \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{\sqrt{1-x^2}}dx$

I get stuck here, using U and V for new functions and using the relationship again doesn't lead me anywhere.

2. ## Re: Calculating intregral

Originally Posted by fkf
I've been trying to calculating the following integral
$\int xsin^{-1}(x)\, dx$

Here's my solution as far as I've been able to calculate. Let
$U=sin^{-1}x \Rightarrow dU=\frac{1}{\sqrt{1-x^2}}dx$
$dV = x \cdot dx \Rightarrow V = \frac{x^2}{2}$
Then
$I=VU-\int V\cdot du=sin^{-1}x \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{\sqrt{1-x^2}}dx$

I get stuck here, using U and V for new functions and using the relationship again doesn't lead me anywhere.
You can see how it works here.

Ckick the show steps tab.

3. ## Re: Calculating intregral

Originally Posted by fkf
I've been trying to calculating the following integral
$\int xsin^{-1}(x)\, dx$

Here's my solution as far as I've been able to calculate. Let
$U=sin^{-1}x \Rightarrow dU=\frac{1}{\sqrt{1-x^2}}dx$
$dV = x \cdot dx \Rightarrow V = \frac{x^2}{2}$
Then
$I=VU-\int V\cdot du=sin^{-1}x \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{\sqrt{1-x^2}}dx$

I get stuck here, using U and V for new functions and using the relationship again doesn't lead me anywhere.
Hi, looking at the new integral that you have to solve, you should notice the denominator suggests a possible solution using a trigonometric substitution. Consider trying:

$x = \sin \theta \Rightarrow dx = \cos \theta d\theta$

And then working out that new integral. You will get something that may require the use of an identity of the form:
$\sin^2 \theta = cos2\theta/2 + 1/2$

or something like that (ill leave the details to you).