Re: Calculating intregral

Quote:

Originally Posted by

**fkf** I've been trying to calculating the following integral

$\displaystyle \int xsin^{-1}(x)\, dx$

Here's my solution as far as I've been able to calculate. Let

$\displaystyle U=sin^{-1}x \Rightarrow dU=\frac{1}{\sqrt{1-x^2}}dx$

$\displaystyle dV = x \cdot dx \Rightarrow V = \frac{x^2}{2}$

Then

$\displaystyle I=VU-\int V\cdot du=sin^{-1}x \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{\sqrt{1-x^2}}dx$

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.

Re: Calculating intregral

Quote:

Originally Posted by

**fkf** I've been trying to calculating the following integral

$\displaystyle \int xsin^{-1}(x)\, dx$

Here's my solution as far as I've been able to calculate. Let

$\displaystyle U=sin^{-1}x \Rightarrow dU=\frac{1}{\sqrt{1-x^2}}dx$

$\displaystyle dV = x \cdot dx \Rightarrow V = \frac{x^2}{2}$

Then

$\displaystyle I=VU-\int V\cdot du=sin^{-1}x \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{\sqrt{1-x^2}}dx$

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:

$\displaystyle 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:

$\displaystyle \sin^2 \theta = cos2\theta/2 + 1/2$

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