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.....
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.....
Hello, jsu03!
You're doing fine . . .
$\displaystyle 1)\;\;\int\sin^{-1}\!x\,dx$
I let: .$\displaystyle \begin{array}{ccccccc} u&=&\sin^{-1}\!x && dv&=& dx \\
du&=&\frac{dx}{\sqrt{1-x^2}} && v&=&x \end{array}$
So: .$\displaystyle \int\sin^{-1}\!x\,dx \;\;=\;\; x\sin^{-1}\!x - \int \frac{x\, dx}{\sqrt{1-x^2}}$ . . . . Good!
The new integral is: .$\displaystyle \int \left(1-x^2\right)^{-\frac{1}{2}}(x\,dx)$
Now let: $\displaystyle u \,=\,1-x^2 $
Got it?