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

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

Vedclass · Accepted Answer

$15\pi$

Vedclass · Answer

(C) We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2(2x)$.Using the identity $\sin^2(2x) = \frac{1 - \cos(4x)}{2}$,we get $\sin^4 x + \cos^4 x = 1 - \frac{1}{2} \left( \frac{1 - \cos(4x)}{2} \right) = 1 - \frac{1}{4} + \frac{1}{4} \cos(4x) = \frac{3}{4} + \frac{1}{4} \cos(4x)$.Now,integrating from $0$ to $20\pi$:$\int_0^{20\pi} (\frac{3}{4} + \frac{1}{4} \cos 4x) dx = [\frac{3}{4}x + \frac{1}{16} \sin 4x]_0^{20\pi}$.Substituting the limits: $(\frac{3}{4} \cdot 20\pi + \frac{1}{16} \sin(80\pi)) - (0 + 0) = 15\pi + 0 = 15\pi$.

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

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