Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 56

$\frac{1}{\pi}\sin^{-1}(x^{1/\pi})\cdot (x^{\frac{1-\pi}{\pi}})$

Given $$y=\int_{-1}^{x^{1 / \pi}} \sin ^{-1} t d t$$ Since \begin{aligned}\frac{dy}{dx}&=\frac{d }{dx}\int_{-1}^{x^{1 / \pi}} \sin ^{-1} t d t\\ &=\sin^{-1}(x^{1/\pi}) \frac{d}{dx}(x^{1/\pi})\\ &= \frac{1}{\pi}\sin^{-1}(x^{1/\pi})\cdot (x^{\frac{1-\pi}{\pi}}) \end{aligned}