int_{-frac{pi}{4}}^{frac{pi}{4}} sin^4 x , dx | vedclass

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Question

(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^4 x \, dx$. Since $f(x) = \sin^4 x$ is an even function,$I = 2 \int_{0}^{\frac{\pi}{4}} \sin^4

Vedclass · Accepted Answer

$\frac{3\pi - 8}{16}$

Vedclass · Answer

(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^4 x \, dx$. Since $f(x) = \sin^4 x$ is an even function,$I = 2 \int_{0}^{\frac{\pi}{4}} \sin^4 x \, dx$.
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we have $\sin^4 x = (\frac{1 - \cos 2x}{2})^2 = \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x)$.
Further,using $\cos^2 2x = \frac{1 + \cos 4x}{2}$,we get $\sin^4 x = \frac{1}{4}(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}) = \frac{1}{4}(\frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x) = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.
Integrating,$I = 2 \int_{0}^{\frac{\pi}{4}} (\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x) \, dx = 2 [\frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x]_{0}^{\frac{\pi}{4}}$.
Evaluating at the limits: $I = 2 [(\frac{3}{8} \cdot \frac{\pi}{4} - \frac{1}{4}\sin \frac{\pi}{2} + \frac{1}{32}\sin \pi) - (0)] = 2 [\frac{3\pi}{32} - \frac{1}{4} + 0] = \frac{3\pi}{16} - \frac{1}{2} = \frac{3\pi - 8}{16}$.

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^4 x \, dx = $

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