int x^2sin^-1(x)

**^_^Engineer_Adam^_^** · August 5th 2007, 09:27 PM

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

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

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

$u = 2x$
$du = 2dx$

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

and my answer wud be:

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

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

**curvature** · August 5th 2007, 10:18 PM

Originally Posted by ^_^Engineer_Adam^_^

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

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

For the 2nd integral let x=sint.

**^_^Engineer_Adam^_^** · August 6th 2007, 02:38 AM

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

how do i simplify to get the corect answer?

**Krizalid** · August 6th 2007, 09:17 AM

Originally Posted by ^_^Engineer_Adam^_^

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

For the second one set $u=\sqrt{1-x^2}$

**curvature** · August 6th 2007, 09:00 PM

Originally Posted by ^_^Engineer_Adam^_^

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

how do i simplify to get the corect answer?

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

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