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Chapter 5: Problem 41

Use a change of variables to evaluate the following definite integrals. $$\int_{0}^{\pi / 2} \sin ^{2} \theta \cos \theta d \theta$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int_{0}^{\pi / 2} \sin ^{2} \theta \cos \theta d \theta.$$ Answer: The evaluated integral is $$\int_{0}^{\pi / 2} \sin ^{2} \theta \cos \theta d \theta = \frac{1}{3}.$$

Step by step solution

01

Identify the substitution

Since the integral contains both sine and cosine functions, it would be beneficial to use a substitution that simplifies the integrand. We can use the substitution \(u = \sin \theta\). Then, the differential, \(du\), is given by: $$du = \frac{d}{d\theta}(\sin \theta) d\theta = \cos \theta d\theta $$ Also, we need to modify the limits of integration according to the new variable \(u\). When \(\theta = 0\), we have \(u = \sin 0 = 0\) and when \(\theta = \frac{\pi}{2}\), we have \(u = \sin (\frac{\pi}{2}) = 1\). The original integral becomes: $$\int_{0}^{1} u^2 \ du$$
02

Evaluate the transformed integral

Now we can easily calculate this simpler integral according to the power rule of integration: $$\int u^2 \ du = \frac{u^3}{3} + C$$ The definite integral will become: $$\int_{0}^{1} u^2 \ du = \left[\frac{u^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$$
03

Present the final result

The evaluated integral of the original problem is: $$\int_{0}^{\pi / 2} \sin ^{2} \theta \cos \theta d \theta = \frac{1}{3}$$

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