# Thread: Integration by Parts #1

1. ## Integration by Parts #1

Can someone help me with this problem? I already started it but got stuck. Thanks.

1. Integral sin^-1 xdx
I let u=sin^-1x dv=dx
du=dx/squareroot 1-x^2 v=x
so integral sin^-1 xdx = x(sin^-1x)-integral x dx/squareroot 1-x^2.....

2. Hello, jsu03!

You're doing fine . . .

$1)\;\;\int\sin^{-1}\!x\,dx$

I let: . $\begin{array}{ccccccc} u&=&\sin^{-1}\!x && dv&=& dx \\
du&=&\frac{dx}{\sqrt{1-x^2}} && v&=&x \end{array}$

So: . $\int\sin^{-1}\!x\,dx \;\;=\;\; x\sin^{-1}\!x - \int \frac{x\, dx}{\sqrt{1-x^2}}$ . . . . Good!

The new integral is: . $\int \left(1-x^2\right)^{-\frac{1}{2}}(x\,dx)$

Now let: $u \,=\,1-x^2$

Got it?

3. if I let u = 1-x^2 then du = -2x and what is dv and v ....I'm confuse...would dv =xdx and v= dx?