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Chapter 7: Problem 8

Evaluate the following integrals. $$\int_{0}^{3 / 2} \frac{d x}{\left(9-x^{2}\right)^{3 / 2}}$$

Short Answer

Expert verified
Using the method of trigonometric substitution, we can find the value of the definite integral $$\int_{0}^{3 / 2} \frac{d x}{\left(9-x^{2}\right)^{3 / 2}}$$ to be \(-\frac{\sqrt{3}}{3}\).

Step by step solution

01

Choose a trigonometric substitution

. Let's choose a trigonometric substitution for the given integral. Since we have \((9-x^2)^{3/2}\) in the denominator, we will use the sine function. Let $$x = 3\sin{\theta}.$$ Now, we need to find the differential, \(dx\), and make the substitution.
02

Find the differential

. Differentiate $$x = 3\sin{\theta}$$ with respect to \(\theta\), which gives: $$dx = 3\cos{\theta} d\theta$$
03

Make the substitution

. Replace \(x\) with \(3\sin{\theta}\) and \(dx\) with \(3\cos{\theta} d\theta\) in the integral: $$\int_{0}^{3 / 2} \frac{d x}{\left(9-x^{2}\right)^{3 / 2}}$$ becomes $$\int \frac{3\cos{\theta}\, d\theta}{(9-9\sin^2{\theta})^{3/2}}$$
04

Simplify the integral

. Now, simplify the integral: $$\int \frac{3\cos{\theta}\, d\theta}{(9-9\sin^2{\theta})^{3/2}} = \int \frac{3\cos{\theta}\, d\theta}{(9(1-\sin^2{\theta})^{3/2}} = \int \frac{3\cos{\theta}\, d\theta}{(9\cos^3{\theta})}$$ Cancel the common factors, and we get: $$\int \frac{1}{3\cos^2{\theta}}\, d\theta$$
05

Find the limits of integration for \(\theta\)

. To change the limits, we now need to find the limits of integration for \(\theta\). Using the substitution \(x = 3\sin{\theta}\): When \(x = 0\), \(\sin{\theta} = 0\), so \(\theta = 0\) When \(x = \frac{3}{2}\), \(\sin{\theta} = \frac{1}{2}\), so \(\theta = \frac{\pi}{6}\)
06

Change the limits of integration and apply the trigonometric identity

. Change the limits of integration and apply the trigonometric identity \(\cos^2{\theta} = 1 - \sin^2{\theta}\): The integral becomes: $$\int_{0}^{\frac{\pi}{6}} \frac{1}{3\left(1-\sin^2{\theta}\right)}\, d\theta $$
07

Integrate

. Now, integrate with respect to \(\theta\): $$\int_{0}^{\frac{\pi}{6}} \frac{1}{3\left(1-\sin^2{\theta}\right)}\, d\theta = \frac{1}{3}\int_{0}^{\frac{\pi}{6}} \frac{d\theta}{1-\sin^2{\theta}}$$ The integral has a standard form: $$\int \frac{d\theta}{1-\sin^2{\theta}} = \int \csc^2{\theta}\, d\theta = -\cot{\theta} + C$$ So, we get: $$\frac{1}{3}(-\cot{\theta} + C)\Big|_{0}^{\frac{\pi}{6}}$$
08

Evaluate the integral at the limits

. Now, evaluate the integral at the given limits of integration: $$\frac{1}{3}(-\cot{\theta} + C)\Big|_{0}^{\frac{\pi}{6}} = \frac{1}{3}(-\cot{\frac{\pi}{6}} + \cot{0}) = \frac{1}{3}(-\sqrt{3} + 0) = -\frac{\sqrt{3}}{3}$$ So, the value of the definite integral $$\int_{0}^{3 / 2} \frac{d x}{\left(9-x^{2}\right)^{3 / 2}}$$ is \(-\frac{\sqrt{3}}{3}\).

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