The value of int_{0}^{20pi} (sin^4 x + cos^4 | vedclass

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Question

(C) We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2} \sin^2(2x)$

Vedclass · Accepted Answer

$15\pi$

Vedclass · Answer

(C) We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2} \sin^2(2x)$.Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$, we get $1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} ight) = 1 - \frac{1}{4} + \frac{1}{4} \cos 4x = \frac{3}{4} + \frac{1}{4} \cos 4x$.Now, the integral $I = \int_{0}^{20\pi} (\frac{3}{4} + \frac{1}{4} \cos 4x) dx$.$I = \int_{0}^{20\pi} \frac{3}{4} dx + \int_{0}^{20\pi} \frac{1}{4} \cos 4x dx$.Since the integral of $\cos(nx)$ over a full period is $0$, $\int_{0}^{20\pi} \frac{1}{4} \cos 4x dx = 0$.Therefore, $I = \frac{3}{4} [x]_{0}^{20\pi} = \frac{3}{4} 	imes 20\pi = 15\pi$.

The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x) dx$ is equal to:

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