Question

whats the integral of arc sin x ??

Answer 1

so far i have:\[x(arcsinx)-\int\limits_{}^{}x \frac{ 1 }{ \sqrt{1-x^2} }dx\]

Answer 2

Looks like you've done most of the hard part. Now take 1-x^2 and integrate by subtitution.

Answer 3

uhmm how can i do that?

Answer 4

there's where i got confused!

Answer 5

trig substitution

x = sin(u)
dx = cos(u) du

Answer 6

So take (1-x^2)=u and take the derivative with respect to x, this becomes

-2x=du/dx which can be rearranged with algebra into:

xdx=-(1/2)du

Substitution back into the original integrand gives:
\[-\int\limits_{}^{}\frac{- du }{ 2\sqrt{u} } = \frac{ 1 }{ 2 }\int\limits_{}^{}\ u^{-1/2} du\]

Answer 7

@dumbcow trig substitution isn't really useful here.