# 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{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 \\ \\ \int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \int^{\frac{1}{2}}_0 u du \\ \\ \int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \frac{u^2}{2} |^{\frac{1}{2}}_0 \\ \\ \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} \\ \\ \int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \frac{\displaystyle \frac{\pi^2}{36}}{2} \\ \\ \int^{\frac{1}{2}}_0 \frac{\sin^{-1}x}{\sqrt{1 - x^2}} dx =& \frac{\pi^2}{72} \end{aligned}

Posted on