Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 82

Find the integral $\displaystyle \int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx$

If we let $u = \sin^{-1} x$, then $\displaystyle du = \frac{1}{\sqrt{1 - x^2}} dx$. When $\displaystyle x = \frac{1}{2}, u = \frac{\pi}{6}$. Therefore,



$
\begin{equation}
\begin{aligned}

\int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \int^{\frac{1}{2}}_0 \sin^{-1} x \cdot \frac{1}{\sqrt{1 - x^2}} dx
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\int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \int^{\frac{1}{2}}_0 u du
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\int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \frac{u^2}{2} |^{\frac{1}{2}}_0
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\int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \frac{\displaystyle \left( \frac{\pi}{6} \right)^2 }{ 2 } - \frac{(0)^2}{2}
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\int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \frac{\displaystyle \frac{\pi^2}{36}}{2}
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\int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \frac{\pi^2}{72}

\end{aligned}
\end{equation}
$