Need help with integral with inverse trig

[kluda06]

i have to evaluate the integral using U*V-∫V*DU

∫sin^-12xdx

i just do not know if i am getting my du right. I did:

u=sin^-12x
du= 2/ square root of (1-(2x)²) ????

dv=dx
v=x

Please let me know if i am correct on this so far.

 

[daon2]

yes, but you missed a dx. should be a simple substitution now