Question

use any method to evaluate the following integral.
int_{0}^{2}\frac{dx}{(16 - x^{2})^{\frac{3}{2}}}
int_{0}^{2}\frac{dx}{(16 - x^{2})^{\frac{3}{2}}}=square
(type an exact answer, using radicals as needed.)

Explanation:

Step1: Use trigonometric substitution

Let $x = 4\sin\theta$, then $dx=4\cos\theta d\theta$. When $x = 0$, $\theta=0$; when $x = 2$, $\sin\theta=\frac{1}{2}$, so $\theta=\frac{\pi}{6}$. And $16 - x^{2}=16-16\sin^{2}\theta = 16\cos^{2}\theta$.

Step2: Rewrite the integral

The integral $\int_{0}^{2}\frac{dx}{(16 - x^{2})^{\frac{3}{2}}}$ becomes $\int_{0}^{\frac{\pi}{6}}\frac{4\cos\theta d\theta}{(16\cos^{2}\theta)^{\frac{3}{2}}}=\int_{0}^{\frac{\pi}{6}}\frac{4\cos\theta d\theta}{64\cos^{3}\theta}=\frac{1}{16}\int_{0}^{\frac{\pi}{6}}\frac{d\theta}{\cos^{2}\theta}$.

Step3: Integrate $\frac{1}{\cos^{2}\theta}$

We know that $\int\frac{d\theta}{\cos^{2}\theta}=\tan\theta + C$. So $\frac{1}{16}\int_{0}^{\frac{\pi}{6}}\frac{d\theta}{\cos^{2}\theta}=\frac{1}{16}[\tan\theta]_{0}^{\frac{\pi}{6}}$.

Step4: Evaluate the definite - integral

$\frac{1}{16}(\tan\frac{\pi}{6}-\tan0)=\frac{1}{16}(\frac{\sqrt{3}}{3}-0)=\frac{\sqrt{3}}{48}$.

Answer:

$\frac{\sqrt{3}}{48}$