int_{0}^{20 pi}(|sin x|+|cos x|)^{2} d x is e | vedclass

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(D) Let $I = \int_{0}^{20 \pi} (|\sin x| + |\cos x|)^2 dx$.Since the function $f(x) = (|\sin x| + |\cos x|)^2$ is periodic with period $\frac{\pi}{2}$

Vedclass · Accepted Answer

$20(\pi+2)$

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(D) Let $I = \int_{0}^{20 \pi} (|\sin x| + |\cos x|)^2 dx$.Since the function $f(x) = (|\sin x| + |\cos x|)^2$ is periodic with period $\frac{\pi}{2}$,we can use the property $\int_{0}^{nT} f(x) dx = n \int_{0}^{T} f(x) dx$.Here,$n = \frac{20\pi}{\pi/2} = 40$ and $T = \frac{\pi}{2}$.Thus,$I = 40 \int_{0}^{\pi/2} (|\sin x| + |\cos x|)^2 dx$.Since $\sin x \ge 0$ and $\cos x \ge 0$ for $x \in [0, \pi/2]$,we have:$I = 40 \int_{0}^{\pi/2} (\sin x + \cos x)^2 dx$.Expanding the square,we get $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x$.So,$I = 40 \int_{0}^{\pi/2} (1 + \sin 2x) dx$.$I = 40 [x - \frac{\cos 2x}{2}]_{0}^{\pi/2}$.$I = 40 [(\frac{\pi}{2} - \frac{\cos \pi}{2}) - (0 - \frac{\cos 0}{2})]$.$I = 40 [(\frac{\pi}{2} + \frac{1}{2}) - (0 - \frac{1}{2})] = 40 [\frac{\pi}{2} + 1] = 20(\pi + 2)$.

$\int_{0}^{20 \pi}(|\sin x|+|\cos x|)^{2} d x$ is equal to.

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