integration of inverse functions?

Question

integral sin-1 sqrt(x) - cos-1sqrt(x) / sin-1 sqrt(x) + cos-1sqrt(x)

-1 stands for inverse of
i got
as denom is pi/2 and cos-1 sqrt(x) = pi/2 - sin-1 sqrt(x)
sin-1 sqrt(x) gets cancelled

but answer given is ( 2(2x-1) / pi sin-1 sqrt(x) )+
(2 sqrt(x- x^2)/pi ) -x + c

please help me.

Answer 1

You're on the right track. You've simplified the fraction corrrectly, and if you carry your algebra just a little further you can see that what you have is at least consistent with the given answer. Break your fraction up into two fractions, and you get

(4/pi) inversesin sqrt x -1

at least you can see where -x +C comes from in the answer.

As for integrating the first term, make the substitution

t = sqrtx

After a bit of algebra it will lead to dx = 2 sqrt x dt, or
dx =2tdt.

so you have to integrate Integral { t inversesine t dt}.

Use integration by parts, with u=inversesin t.