Question

How do I do the integral of sin^(-1)(x)dx?

Answer 1

\[\int\limits_{}^{}\sin^{-1}x dx\]

Answer 2

i am going to guess parts

Answer 3

matter of fact i am almost sure it is parts
\[\int u dv =uv-\int vdu\] with
\[u=\sin^{-1}(x), du =\frac{1}{\sqrt{1-x^2}}, dv=1dx, v =x\]

Answer 4

same sort of trick as
\[\int \ln(x)dx\]

Answer 5

you get
\[x\sin^{-1}(x)-\int \frac{x}{\sqrt{1-x^2}}dx\] and the second integral is a relatively easy \(u\) sub with \(u=1-x^2, du=-2xdx\) etc

Answer 6

Oh I see where the parts are now, thank you.