# Chapter 7 - Techniques of Integration - 7.3 Trigonometric Substitution - 7.3 Exercises - Page 531: 15

$$\displaystyle{\int_0^{a}x^2\sqrt{a^2-x^2}dx=\frac{a^4\pi }{16}}\\$$

#### Work Step by Step

$\displaystyle{I=\int_0^{a}x^2\sqrt{a^2-x^2}dx}\\$ $\displaystyle \left[\begin{array}{ll} x=a\sin\theta & 9 x^2=a^2\sin^2\theta \\ & \\ \frac{dx}{d\theta}=a\cos\theta & dx=a\cos\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int_0^{\frac{\pi}{2}}a^2\sin^2\theta\sqrt{a^2-a^2\sin^2\theta}\times a\cos\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{2}}a^2\sin^2\theta\sqrt {a^2\left(1-\sin^2\theta\right)}\times a\cos\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{2}}a^2\sin^2\theta\times a\cos\theta\times a\cos\theta\ d\theta}\\ \displaystyle{I=a^4\int_0^{\frac{\pi}{2}}\cos^2\theta\sin^2\theta\ d\theta}\\ \displaystyle{I=a^4\int_0^{\frac{\pi}{2}}\frac{1}{4}\times4\cos^2\theta\sin^2\theta\ d\theta}\\ \displaystyle{I=\frac{a^4}{4}\int_0^{\frac{\pi}{2}}\sin^22\theta\ d\theta}\\ \displaystyle{I=\frac{a^4}{4}\int_0^{\frac{\pi}{2}}\frac{1}{4}\left(1-\cos4\theta\right)\ d\theta}\\ \displaystyle{I=\frac{a^4}{8}\int_0^{\frac{\pi}{2}}1-\cos4\theta\ d\theta}\\ \displaystyle{I=\frac{a^4}{8}\left[\theta-\frac{1}{4}\sin4\theta\right]_0^{\frac{\pi}{2}}}\\ \displaystyle{I=\frac{a^4\pi }{16}-\frac{a^4}{32}\sin\left(4\times\frac{\pi }{2}\right)-0-0}\\ \displaystyle{I=\frac{a^4\pi }{16}-0}\\ \displaystyle{I=\frac{a^4\pi }{16}}\\$

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