Question

use any method to evaluate the following integral.
int_{0}^{2sqrt{2}} \frac{dx}{(16 - x^{2})^{\frac{3}{2}}}
int_{0}^{2sqrt{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 $x = 0$, $\sin\theta=0$, so $\theta = 0$. When $x = 2\sqrt{2}$, $\sin\theta=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}$, so $\theta=\frac{\pi}{4}$. Also, $16 - x^{2}=16-16\sin^{2}\theta = 16\cos^{2}\theta$.

Step2: Rewrite the integral

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

Step3: Integrate the new - form integral

Since $\frac{1}{\cos^{2}\theta}=\sec^{2}\theta$ and $\int\sec^{2}\theta d\theta=\tan\theta + C$, then $\frac{1}{16}\int_{0}^{\frac{\pi}{4}}\frac{d\theta}{\cos^{2}\theta}=\frac{1}{16}[\tan\theta]_{0}^{\frac{\pi}{4}}$.

Step4: Evaluate the definite integral

$\frac{1}{16}(\tan\frac{\pi}{4}-\tan0)=\frac{1}{16}(1 - 0)=\frac{1}{16}$.

Answer:

$\frac{1}{16}$