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Chapter 5: Problem 60

Integrals with \(\sin ^{2} x\) and \(\cos ^{2} x\) Evaluate the following integrals. $$\int_{0}^{\pi / 2} \sin ^{4} \theta d \theta$$

Short Answer

Expert verified
Question: Evaluate the following definite integral: $$\int_{0}^{\pi/2} \sin^4 \theta d\theta$$ Answer: The definite integral evaluates to $$\int_{0}^{\pi/2} \sin^4 \theta d\theta = \frac{3\pi}{16}$$

Step by step solution

01

Use trigonometric identity

We have to evaluate the definite integral of \(\sin^4 \theta\). We can rewrite this using the double angle trigonometric identity: $$\sin^2 \theta = \frac{1 - \cos(2 \theta)}{2}$$ Now, we want to replace \(\sin^4 \theta\) with an expression in terms of \(\cos\): $$\sin^4 \theta = \left(\frac{1 - \cos(2 \theta)}{2}\right)^2 = \frac{1}{4} (1 - 2\cos(2\theta) + \cos^2(2\theta))$$
02

Use trigonometric identity again

We still have a power 2 in the expression, so use the same identity again to deal with \(\cos^2(2\theta)\): $$\cos^2(2\theta) = \frac{1 + \cos(4 \theta)}{2}$$ Now substitute this in our previous expression: $$\sin^4 \theta = \frac{1}{4} (1 - 2\cos(2\theta) + \frac{1}{2} (1 + \cos(4\theta)))$$
03

Simplify the expression

Combine the terms to simplify the expression: $$\sin^4 \theta = \frac{1}{4} (1 - 2\cos(2\theta) + \frac{1}{2} + \frac{1}{2} \cos(4\theta)) = \frac{1}{8} (3 - 4\cos(2\theta) + \cos(4\theta))$$
04

Evaluate the definite integral

Now, we evaluate the definite integral of this function from \(0\) to \(\pi/2\): $$\int_{0}^{\pi/2} \sin^4 \theta d\theta = \frac{1}{8} \int_{0}^{\pi/2} (3 - 4\cos(2\theta) + \cos(4\theta)) d\theta$$ Now, integrate each term seperately: $$\frac{1}{8} \left[ 3\theta - 2\sin(2\theta) + \frac{1}{4}\sin(4\theta)\right]_{0}^{\pi/2}$$
05

Apply limits and compute the result

Apply the limits of integration and compute the result: $$\frac{1}{8} \left( 3\frac{\pi}{2} - 2\sin(\pi) + \frac{1}{4}\sin(2\pi) - (3\cdot 0 - 2\sin(0) + \frac{1}{4}\sin(0)) \right) = \frac{1}{8} \cdot 3\frac{\pi}{2} = \frac{3\pi}{16}$$ Therefore, the result of the definite integral is: $$\int_{0}^{\pi/2} \sin^4 \theta d\theta = \frac{3\pi}{16}$$

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