Evaluating the Integral of (arcsinx)^2dx using Integration by Parts

[Vash]

Homework Statement

elvaluate this integral (arcsinx)^2dx

Homework Equations

(arcsinx)^2dx

The Attempt at a Solution

integration by parts, let u= arcsinx and make y=arcsinx for easier integration. Once i plug it into the parts equation it turns into a mess. Any help would be superb.

 

[G01]

This should be solvable by using substitution and integration by parts one after the other.

HINT: Let sin u = x.

 

[Vash]

That's what I've been doing and it still doesn't work out for me...oh well, I'll keep trying.

 

[rocomath]

\int(\sin^{-1}x)^2dx

t=\sin^{-1}x
dt=\frac{dx}{\sqrt{1-x^2}}

x=\sin t
dt=\frac{dx}{\sqrt{1-\sin^2 t}}

Continue simplifying and see what you can get.

 

[Vash]

im getting xarcsinx+sqrt(1-x^2)

 

[rocomath]

im getting xarcsinx+sqrt(1-x^2)

Final answer? Take the derivative and see if you get your Integral.

 

[Vash]

Im not sure how to do it your way so I just set it up by integration by parts. u=arcsinx, du=1/(sqrt(1-x^2)) dv=dx v=x. When I do it i get my answer above.

 

[rocomath]

After simplifying, the Integral becomes ...

\int t^2 \cos t dt