Question

$\int \frac{dx}{(64 - x^2)^{3/2}}$

Explanation:

Step1: Choose trigonometric substitution

Let \( x = 8\sin\theta \), then \( dx = 8\cos\theta d\theta \). Also, \( 64 - x^{2}=64 - 64\sin^{2}\theta = 64\cos^{2}\theta \), so \( (64 - x^{2})^{3/2}=(64\cos^{2}\theta)^{3/2}=512\cos^{3}\theta \).

Step2: Substitute into the integral

Substitute \( x \) and \( dx \) into the integral:
\[

$$\begin{align*} \int\frac{dx}{(64 - x^{2})^{3/2}}&=\int\frac{8\cos\theta d\theta}{512\cos^{3}\theta}\\ &=\frac{1}{64}\int\frac{d\theta}{\cos^{2}\theta}\\ &=\frac{1}{64}\int\sec^{2}\theta d\theta \end{align*}$$

\]

Step3: Integrate with respect to \(\theta\)

We know that the integral of \( \sec^{2}\theta \) with respect to \( \theta \) is \( \tan\theta + C \), so:
\[
\frac{1}{64}\int\sec^{2}\theta d\theta=\frac{1}{64}\tan\theta + C
\]

Step4: Convert back to \( x \)

Since \( x = 8\sin\theta \), we have \( \sin\theta=\frac{x}{8} \). Using the right triangle with opposite side \( x \), hypotenuse \( 8 \), the adjacent side is \( \sqrt{64 - x^{2}} \). Then \( \tan\theta=\frac{x}{\sqrt{64 - x^{2}}} \).
Substitute \( \tan\theta \) back into the expression:
\[
\frac{1}{64}\tan\theta + C=\frac{x}{64\sqrt{64 - x^{2}}}+C
\]

Answer:

\(\frac{x}{64\sqrt{64 - x^{2}}}+C\)