Question

Evaluate the integral

Answer 1

Substitute $y=\sin^{-1}\left(\frac x4\right)$, so that

$dy = \dfrac{dx}{4\sqrt{1-\left(\frac x4\right)^2}} = \dfrac{dx}{\sqrt{16 - x^2}}$

Then the integral is

$\displaystyle \int \frac{\left(\sin^{-1}\left(\frac x4\right)\right)^2}{\sqrt{16-x^2}} \,dx = \int y^2 \, dy$

and by the power rule,

$\displaystyle \int y^2 \, dy = \frac13 y^3 + C$

so that the original integral is

$\displaystyle \int \frac{\left(\sin^{-1}\left(\frac x4\right)\right)^2}{\sqrt{16-x^2}} \,dx = \boxed{\frac13 \left(\sin^{-1}\left(\frac x4\right)\right)^3 + C}$