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Chapter 5: Problem 51

Use a change of variables to evaluate the following definite integrals. $$\int_{1 / 3}^{1 / \sqrt{3}} \frac{4}{9 x^{2}+1} d x$$

Short Answer

Expert verified
When evaluating the definite integral $$\int_{1 / 3}^{1 / \sqrt{3}} \frac{4}{9 x^{2}+1} dx$$ the result is $$\frac{\pi}{9}$$.

Step by step solution

01

Choose an appropriate trigonometric substitution

Since the integral has the form: $$\frac{1}{ax^2 + b}$$ where a = 9 and b = 1, an appropriate substitution would be $$x = \frac{1}{3} \tan(\theta)$$ Then, the differential will be: $$dx = \frac{1}{3} \sec^2(\theta) d\theta$$
02

Change the limits of integration

We'll convert the limits of integration from x-values to 胃-values. When x = 1/3: $$\frac{1}{3} = \frac{1}{3} \tan(\theta_1) \Rightarrow \theta_1 = \arctan(1) = \frac{\pi}{4}$$ When x = 1/鈭�3: $$\frac{1}{\sqrt{3}} = \frac{1}{3} \tan(\theta_2) \Rightarrow \theta_2 = \arctan(\sqrt{3}) = \frac{\pi}{3}$$
03

Rewrite the integral using the new variable

Using the substitution and the new limits of integration, the integral becomes: $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{4}{9 (\frac{1}{3}\tan(\theta))^2 + 1} \cdot \frac{1}{3} \sec^2(\theta) d\theta$$ Now, simplify the integral: $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{4}{9 (\tan^2(\theta)) + 1} \sec^2(\theta) d\theta$$ Recall that $$1 + \tan^2(\theta) = \sec^2(\theta)$$, so: $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{4}{9\sec^2(\theta)} \sec^2(\theta) d\theta$$
04

Evaluate the integral

Now, the integral is easy to evaluate: $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{4}{9} d\theta = \frac{4}{9} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} d\theta$$ $$ = \frac{4}{9} [\theta]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{4}{9} (\frac{\pi}{3} - \frac{\pi}{4})$$ $$ = \frac{4}{9} (\frac{\pi}{12}) = \frac{\pi}{9}$$ So, the definite integral: $$\int_{1 / 3}^{1 / \sqrt{3}} \frac{4}{9 x^{2}+1} d x = \frac{\pi}{9}$$

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