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

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• Aug 5th 2007, 09:27 PM
^_^Engineer_Adam^_^
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

and my answer wud be:

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

but the answer at the back of book is:
$\displaystyle \frac{1}{3}x^3\sin^{-1}{x} + \frac{1}{9}(x^2+2)\sqrt{1-x^2} + C$
• Aug 5th 2007, 10:18 PM
curvature
Quote:

Originally Posted by ^_^Engineer_Adam^_^
$\displaystyle \int x^2\sin^{-1}{x}$
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.
• Aug 6th 2007, 02:38 AM
^_^Engineer_Adam^_^
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?
• Aug 6th 2007, 09:17 AM
Krizalid
Quote:

Originally Posted by ^_^Engineer_Adam^_^
$\displaystyle \frac{x^3}{3}\sin^{-1}{x} - \int\frac{x^3}{3}\frac{1}{(\sqrt{1-x^2})}$

For the second one set $\displaystyle u=\sqrt{1-x^2}$
• Aug 6th 2007, 09:00 PM
curvature
Quote:

Originally Posted by ^_^Engineer_Adam^_^
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?

The correct answer is
$\displaystyle \frac{1}{3}x^3\sin^{-1}x + \frac{1}{3}\sqrt{1-x^2} - \frac{1}{9}(1-x^2)^{\frac{3}{2}}$
where