1. ## int x^2sin^-1(x)

$\displaystyle \int x^2\sin^{-1}{x}$
let dv = x^2
v = x^3/3
let u = sin^-1{x}
$\displaystyle du = \frac{1}{(\sqrt{1 + x^2}} dx$

$\displaystyle \frac{x^3}{3}\sin^{-1}{x} - \int\frac{x^3}{3}\frac{1}{(\sqrt{1-x^2})}$

for the 2nd integral:
$\displaystyle dv = \frac{x}{\sqrt{1-x^2}}dx$
$\displaystyle v = - \sqrt{1-x^2}$

$\displaystyle u = 2x$
$\displaystyle du = 2dx$

$\displaystyle -x^2\sqrt{1-x^2} - \int -2x\sqrt{1-x^2}dx$
integrat

$\displaystyle \frac{x^3}{3}\sin^{-1}{x} +\frac{1}{3}x^2\sqrt{1-x^2} +\frac{2}{9}(1-x^2)^{\frac{3}{2}}$

but the answer at the back of book is:
$\displaystyle \frac{1}{3}x^3\sin^{-1}{x} + \frac{1}{9}(x^2+2)\sqrt{1-x^2} + C$

$\displaystyle \int x^2\sin^{-1}{x}$
let dv = x^2
v = x^3/3
let u = sin^-1{x}
$\displaystyle du = \frac{1}{(\sqrt{1 + x^2}} dx$

$\displaystyle \frac{x^3}{3}\sin^{-1}{x} - \int\frac{x^3}{3}\frac{1}{(\sqrt{1-x^2})}dx$
......
for the 2nd integral
For the 2nd integral let x=sint.

3. using trigo sub
i will get a value of
$\displaystyle \frac{1}{3}\sin^{-1}x + \frac{1}{3}\sqrt{1-x^2} - \frac{1}{9}(1-x^2)^{\frac{3}{2}}$

how do i simplify to get the corect answer?

$\displaystyle \frac{x^3}{3}\sin^{-1}{x} - \int\frac{x^3}{3}\frac{1}{(\sqrt{1-x^2})}$
For the second one set $\displaystyle u=\sqrt{1-x^2}$

$\displaystyle \frac{1}{3}\sin^{-1}x + \frac{1}{3}\sqrt{1-x^2} - \frac{1}{9}(1-x^2)^{\frac{3}{2}}$
$\displaystyle \frac{1}{3}x^3\sin^{-1}x + \frac{1}{3}\sqrt{1-x^2} - \frac{1}{9}(1-x^2)^{\frac{3}{2}}$