The value of int_a^{a + (pi /2)} (sin^4 x + c | vedclass

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(C) Let $I = \int_a^{a + (\pi /2)} (\sin^4 x + \cos^4 x) \, dx$. Since the integrand $f(x) = \sin^4 x + \cos^4 x$ is a periodic function with period $

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

Vedclass · Answer

(C) Let $I = \int_a^{a + (\pi /2)} (\sin^4 x + \cos^4 x) \, dx$. Since the integrand $f(x) = \sin^4 x + \cos^4 x$ is a periodic function with period $\frac{\pi}{2}$,the integral over any interval of length $\frac{\pi}{2}$ is constant and equal to the integral over $[0, \pi/2]$. Thus,$I = \int_0^{\pi/2} (\sin^4 x + \cos^4 x) \, dx$. Using the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$. Therefore,$I = \int_0^{\pi/2} (1 - \frac{1}{2}\sin^2(2x)) \, dx = \int_0^{\pi/2} 1 \, dx - \frac{1}{2} \int_0^{\pi/2} \sin^2(2x) \, dx$. $I = [x]_0^{\pi/2} - \frac{1}{2} \int_0^{\pi/2} \frac{1 - \cos(4x)}{2} \, dx = \frac{\pi}{2} - \frac{1}{4} [x - \frac{\sin(4x)}{4}]_0^{\pi/2}$. $I = \frac{\pi}{2} - \frac{1}{4} (\frac{\pi}{2} - 0) = \frac{\pi}{2} - \frac{\pi}{8} = \frac{3\pi}{8}$.

The value of $\int_a^{a + (\pi /2)} (\sin^4 x + \cos^4 x) \, dx$ is

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