evaluate the following integral using trigonometric substitution.\n int_{0}^{2sqrt{2}}\frac{x^{2}}{sqrt{16…

evaluate the following integral using trigonometric substitution.\n int_{0}^{2sqrt{2}}\frac{x^{2}}{sqrt{16 - x^{2}}}dx \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 4\tan\theta)\nb. (x = 4sec\theta)\nc. (x = 4sin\theta)\nrewrite the given integral using this substitution.\n int_{0}^{2sqrt{2}}\frac{x^{2}}{sqrt{16 - x^{2}}}dx=int_{0}^{square}(square)d\theta \n(type exact answers.)

evaluate the following integral using trigonometric substitution.\n int_{0}^{2sqrt{2}}\frac{x^{2}}{sqrt{16 - x^{2}}}dx \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 4\tan\theta)\nb. (x = 4sec\theta)\nc. (x = 4sin\theta)\nrewrite the given integral using this substitution.\n int_{0}^{2sqrt{2}}\frac{x^{2}}{sqrt{16 - x^{2}}}dx=int_{0}^{square}(square)d\theta \n(type exact answers.)

Answer

Explanation:

Step1: Substitute $x = 4\sin\theta$ and find $dx$

If $x = 4\sin\theta$, then $dx=4\cos\theta d\theta$. Also, when $x = 0$, $0 = 4\sin\theta$ gives $\theta=0$; when $x = 2\sqrt{2}$, $2\sqrt{2}=4\sin\theta$ gives $\sin\theta=\frac{\sqrt{2}}{2}$, so $\theta=\frac{\pi}{4}$. And $\sqrt{16 - x^{2}}=\sqrt{16-16\sin^{2}\theta}=4\cos\theta$.

Step2: Rewrite the integral

Substitute $x$, $dx$ and $\sqrt{16 - x^{2}}$ into the original integral: [ \begin{align*} \int_{0}^{2\sqrt{2}}\frac{x^{2}}{\sqrt{16 - x^{2}}}dx&=\int_{0}^{\frac{\pi}{4}}\frac{(4\sin\theta)^{2}}{4\cos\theta}\cdot4\cos\theta d\theta\ &=\int_{0}^{\frac{\pi}{4}}16\sin^{2}\theta d\theta \end{align*} ]

Answer:

$\int_{0}^{\frac{\pi}{4}}16\sin^{2}\theta d\theta$