Question

Evaluate (medal given for steps shown ^_^)
\[ \huge \int_{0}^{1/4} \frac{sin^{-1}(2x)}{\sqrt{1-4x^2}} dx \]

Answer 1

Okay, so we need to do a u-substitution right? So let:
\[u=\arcsin(2x) \implies du = \frac{2 dx}{\sqrt{1-4x^2}} \implies \frac{du}{2}=\frac{dx}{\sqrt{1-4x^2}}\]
Therefore the integral becomes:
\[\int\limits_{u_1}^{u_2} \frac{u}{2}du\]
Evaluate your bounds:
arcsin(2(1/4))=arcsin(1/2)=pi/6 (assuming typical branch cut on arcsin from -pi/2 to pi/2)
and arcsin(2(0))=0 So we have:
\[\int\limits_0^{\pi/6}\frac{u}{2}du=\left[ \frac{u^2}{4}\right]_0^{\pi/6}=\frac{\pi^2}{144}\]

Answer 2

Verification:
http://www.wolframalpha.com/input/?i=integral+of+arcsin%282x%29%2F%28sqrt%281-4x%5E2%29%29dx%2Cx%2C0%2C1%2F4

Answer 3

Excellent, ty