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Chapter 5: Problem 57

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

Short Answer

Expert verified
Question: Evaluate the definite integral \(\int_{-\pi / 4}^{\pi / 4} \sin^2(2\theta)d\theta\). Answer: The definite integral evaluates to \(\frac{\pi}{2}\).

Step by step solution

01

Utilize the double-angle formula for sine

Recall the double-angle formula for sine: \(\sin(2x) = 2\sin(x)\cos(x)\). We have \(\sin^2(2\theta)\) in our integral, so we can rewrite our integral as $$\int_{-\pi / 4}^{\pi / 4} \left(2\sin\theta\cos\theta\right)^{2} d\theta$$
02

Simplify the integrand

Continue to simplify the expression inside the integral: $$\int_{-\pi / 4}^{\pi / 4} 4\sin^2\theta\cos^2\theta d\theta$$
03

Apply the trigonometric identity

We can use the trigonometric identity: \(\sin^2(x) = 1-\cos^2(x)\) to rewrite the integrand in terms of a single trigonometric function: $$\int_{-\pi / 4}^{\pi / 4} 4\left(1-\cos^2\theta\right)\cos^2\theta d\theta$$ Now simplify: $$\int_{-\pi / 4}^{\pi / 4} 4\cos^2\theta - 4\cos^4\theta d\theta$$
04

Applying the power-reduction formulas

Begin by applying the power-reduction formula for \(\cos^2\theta\): $$\int_{-\pi / 4}^{\pi / 4} 4\left(\frac{1+\cos(2\theta)}{2}\right) - 4\left(\frac{1+\cos(4\theta)}{2}\right)^2 d\theta$$ Which simplifies to: $$\int_{-\pi/4}^{\pi/4} 2 + 2\cos(2\theta)-\frac{1}{4} -\frac{1}{2}\cos(4\theta) + \frac{1}{4}\cos^2(4\theta) d\theta$$
05

Evaluate the integral

Evaluate the integral by finding the antiderivatives of each term: $$F(\theta) = 2\theta + \sin(2\theta) - \frac{1}{4}\theta - \frac{1}{8}\sin(4\theta) + \frac{1}{32}\sin(8\theta) + C$$ Now apply the Fundamental Theorem of Calculus: $$\int_{-\pi/4}^{\pi/4} 2 + 2\cos(2\theta)-\frac{1}{4} -\frac{1}{2}\cos(4\theta) + \frac{1}{4}\cos^2(4\theta) d\theta = F(\frac{\pi}{4}) - F(-\frac{\pi}{4})$$
06

Calculate the result

Calculate the value of the definite integral using the antiderivatives and the limits of integration: $$F(\frac{\pi}{4}) - F(-\frac{\pi}{4}) = \left[2\cdot\frac{\pi}{4} + \sin(\frac{\pi}{2}) - \frac{1}{4}\cdot\frac{\pi}{4} - \frac{1}{8}\sin(\pi) + \frac{1}{32}\sin(2\pi) \right] - \left[2\cdot(-\frac{\pi}{4}) + \sin(-\frac{\pi}{2}) - \frac{1}{4}\cdot(-\frac{\pi}{4}) - \frac{1}{8}\sin(0) + \frac{1}{32}\sin(0) \right]$$ After simplifying, we find the result: $$\int_{-\pi/4}^{\pi/4} \sin^2(2\theta)d\theta = \frac{\pi}{2}$$

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Most popular questions from this chapter

Find the area of the region \(R\) bounded by the graph of \(f\) and the \(x\) -axis on the given interval. Graph \(f\) and show the region \(R\) $$f(x)=2-|x| ;[-2,4]$$

Consider the function \(f\) and the points \(a, b,\) and \(c\) a. Find the area function \(A(x)=\int_{a}^{x} f(t) d t\) using the Fundamental Theorem. b. Graph \(f\) and \(A\) c. Evaluate \(A(b)\) and \(A(c)\) and interpret the results using the graphs of part \((b)\) $$f(x)=\cos \pi x ; a=0, b=\frac{1}{2}, c=1$$

Find the area of the following regions. The region bounded by the graph of \(f(x)=\frac{x}{\sqrt{x^{2}-9}}\) and the \(x\) -axis between \(x=4\) and \(x=5\).

Consider the function g. which is given in terms of a definite integral with a variable upper limit. a. Graph the integrand. b. Calculate \(g^{\prime}(x)\) c. Graph g, showing all your work and reasoning. $$g(x)=\int_{0}^{x}\left(t^{2}+1\right) d t$$

Consider the function g. which is given in terms of a definite integral with a variable upper limit. a. Graph the integrand. b. Calculate \(g^{\prime}(x)\) c. Graph g, showing all your work and reasoning. $$g(x)=\int_{0}^{x} \sin \left(\pi t^{2}\right) d t \text { (a Fresnel integral) }$$
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