Question-216162

Question Number 216162 by Spillover last updated on 28/Jan/25

Answered by MrGaster last updated on 02/Feb/25

Let x = \frac{π}{3} - t \Rightarrow d x = - d t

\int_{0}^{\frac{π}{3}} \sqrt{\sin^{3} x \sin (\frac{π}{3} - x)} d x \int_{\frac{π}{3}}^{0} \sqrt{\sin^{3} (\frac{π}{3} - t) \sin t} (- d t) = \int_{0}^{\frac{π}{3}} \sqrt{\sin^{3} (\frac{π}{3} - t) \sin t d t}

\int_{0}^{\frac{π}{3}} \sqrt{\sin^{3} x \sin (\frac{π}{3} - x)} d x = \int_{0}^{\frac{π}{3}} \sqrt{\sin^{3} (\frac{π}{3} - x) \sin x d x}

\int_{0}^{\frac{π}{3}} \sqrt{\sin^{3} x \sin (\frac{π}{3} - x)} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} (\sqrt{\sin^{3} x \sin (\frac{π}{3} - x)} + \sqrt{\sin^{3} (\frac{π}{3} - x) \sin x}) d x

\int_{0}^{\frac{π}{3}} \sqrt{\sin^{3} x \sin (\frac{π}{3} - x)} d x = \frac{1}{2} \int_{0}^{\frac{π}{3}} \sqrt{\sin^{3} (\frac{π}{3} / 2)} d x = \frac{π}{16}

so :

\int_{0}^{\frac{π}{3}} \sqrt{\sin^{3} x \sin (\frac{π}{3} - x)} d x = \frac{π}{16}

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