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Chapter 7: Problem 23

Evaluate the following integrals. $$\int \frac{d x}{\sqrt{36-x^{2}}}$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int \frac{dx}{\sqrt{36-x^{2}}}\). Answer: The evaluated integral is \(\int \frac{dx}{\sqrt{36-x^{2}}} = \arcsin\left(\frac{x}{6}\right) + C\).

Step by step solution

01

Identify the substitution

Using the trigonometric substitution, let \(x = 6\sin{\theta}\). Now, we need to find \(dx\), the differential of \(x\) with respect to \(\theta\). Differentiate both sides with respect to \(\theta\) to get, $$\frac{dx}{d\theta} = 6\cos{\theta}$$ Now, multiply both sides by \(d\theta\) to get the \(dx\). $$dx = 6\cos{\theta} d\theta$$
02

Substitute and simplify the integral

Replace \(x\) with the substitution, and \(dx\) with the expression we found earlier: $$\int \frac{1}{\sqrt{36- (6\sin\theta)^2}} \cdot 6\cos{\theta} d\theta$$ Simplify the expression inside the square root: $$\int \frac{6\cos\theta}{\sqrt{36- 36\sin^2\theta}} d\theta$$ Now, recall that \(\sin^2\theta + \cos^2\theta = 1\). We can rewrite the expression under the square root as \(36\cos^2\theta\): $$\int \frac{6\cos\theta}{\sqrt{36\cos^2\theta}}d\theta$$ Simplify by taking the square root and cancelling the common term: $$\int \frac{6\cos\theta}{6\cos\theta} d\theta$$ This simplifies to: $$\int 1 d\theta$$
03

Integrate with respect to \(\theta\)

Integrating \(1\) with respect to \(\theta\) gives: $$\int 1d\theta = \theta + C$$
04

Substitute back for \(x\)

Now, we need to express the answer in terms of \(x\) instead of \(\theta\). Recall our substitution: \(x = 6\sin\theta\). Solve for \(\theta\) to make the substitution: $$\sin\theta = \frac{x}{6}$$ $$\theta = \arcsin\left(\frac{x}{6}\right)$$ Replace \(\theta\) in our antiderivative with our expression in terms of \(x\): $$\arcsin\left(\frac{x}{6}\right) + C$$
05

Final Answer

The evaluated integral is: $$\int \frac{dx}{\sqrt{36-x^{2}}} = \arcsin\left(\frac{x}{6}\right) + C$$

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