pls help me in solving for : integral Arcsin(x) / x^2 dx?

Question

is it solved by substituting or by parts?
[Arcsin(x) mean sin inverse of ( x) ]

Answer 1

I believe Integration by Parts would be your best bet here, since (a) it's straightforward to differentiate Arcsin(x), and (b) it's not hard to integrate x^(-2).

Let u = Arcsin(x) and let dv = x^(-2) dx
Thus du = 1/sqrt(1-x^2) dx and v = -1/x

Set up the IBP formula:

Int [Arcsin(x)/x^2 dx] = uv - Int [v du]
= Arcsin(x)/x - Int [-1/(x sqrt(1-x^2)) dx]
= Arcsin(x)/x + Int [ x^(-1) (1-x^2)^(-1/2) dx]
... (by substitution*)
= Arcsin(x)/x + ln(x) - ln (sqrt (1-x^2) + 1)

Be SURE to visit the Integrator at Wolfram.com (see link below)!

* See also worked-out example #6 at:
http://mathforum.org/mathtools/cell/c,15.14.6,ALL,ALL/

Answer 2

∫Arcsin(x) / x^2 dx

let Arcsin(x) = t
=>x=sin(t)
=>dx=cos(t)dt

=>∫Arcsin(x) / x^2 dx = ∫{t/[sin(t)]^2} cos(t)dt

=>∫{t/[sin(t)]^2} cos(t)dt= ∫t cot(t)cosec(t)dt

using integration by parts,

=>∫t cot(t)cosec(t)dt=-tcosec(t) + ∫cosec(t)dt

= -tcosec(t) + log|cosec(t) + cot(t)| + C

substitute t=Arcsin(x)

=>-Arcsin(x) cosec[Arcsin(x)] + log |cosec(Arcsin(x)) +
cot(Arcsin(x))| + C

Results used:
1)∫cosec(x)cot(x)dx=-cosec(x) + C
2)∫cosex(x)dx=log|cosec(x) + cot(x)| + C

Answer 3

we shall solve it by substitution
let Arcsinx=y
so x=siny
differentiate
dx=cosydy
now we have
integralArcsinx/x^2dx=int.ycosy/(siny)^2dy
=int.ycoty cosecydy
integrate by parts taking y as the first function &
cotycosecy as second function
=y.(-cosecy)-int.1.(-cosecy)dy
=-ycosecy+intcosecydy
=-ycosecy+loglcosecy-cotyl+c
=-y/siny+logl1/siny-cosy/sinyl+c
=-y/siny+logl{1-[1-(siny)^2]^1/2}/sinyl+c
now put back siny=x
&y=Arcsinx
=-Arcsinx/x+logl{1-[1-x^2]^1/2}/xl+c