Question

Evaluate the integral.

Answer 1

\[\int\limits \sin ^{-1}xdx\]

Answer 2

by parts :
\(u = \sin^{-1}x\)
\(v = x\)

Answer 3

\(\large \mathbb{\int uv' = uv - \int u'v}\)

Answer 4

\(\large \mathbb{\int \sin^{-1}x . 1 dx = \sin^{-1}x . x - \int (\sin^{-1}x)' . x ~dx} \)

Answer 5

\(\large \mathbb{~~~~~~~~ = x \sin^{-1}x - \int \frac{1}{\sqrt{1-x^2}} . x~ dx}\)

Answer 6

evaluating that is bit easy...

Answer 7

do i have to integrate 1/sqrt1-x^2 x dx

Answer 8

yess
use substitution..

Answer 9

by parts -- as the name says is really by parts and pieces lol... u need to evaluate the right side integral again... everytime u do by parts, it gives u a simpler integral to evaluate...

Answer 10

can i put u=1/sqrt1-x^2 and dv=x

Answer 11

\(\large \mathbb{~~~~~~~~ = x \sin^{-1}x - \int \frac{1}{\sqrt{1-x^2}} . x~ dx}\)

Answer 12

put u = 1-x^2
du = -2x dx

-du/2 = xdx

Answer 13

\(\large \mathbb{~~~~~~~~ = x \sin^{-1}x - \int \frac{1}{\sqrt{u}} .(-du/2)}\)

Answer 14

\(\large \mathbb{~~~~~~~~ = x \sin^{-1}x + \frac{1}{2} \int u^{-1/2} du}\)

Answer 15

\(\large \mathbb{~~~~~~~~ = x \sin^{-1}x + \frac{1}{2} \frac{u^{-1/2+1}}{-1/2+1}}\)

Answer 16

\(\large \mathbb{~~~~~~~~ = x \sin^{-1}x + \frac{1}{2} \frac{u^{1/2}}{1/2}}\)

Answer 17

\(\large \mathbb{~~~~~~~~ = x \sin^{-1}x + \sqrt{u}}\)

\(\large \mathbb{~~~~~~~~ = x \sin^{-1}x + \sqrt{1-x^2}}\)

Answer 18

add the integration constant \(+C\)

Answer 19

see if that makes more or less sense..

Answer 20

yes it does thank ypu