frac{3}{25} int_0^{25 pi} sqrt{|cos x - cos^3 | vedclass

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(B) Let $I = \frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx$. Since the function $f(x) = \sqrt{|\cos x - \cos^3 x|}$ has a period of $\

Vedclass · Accepted Answer

$4$

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(B) Let $I = \frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx$. Since the function $f(x) = \sqrt{|\cos x - \cos^3 x|}$ has a period of $\pi$, we can write: $I = \frac{3}{25} 	imes 25 \int_0^{\pi} \sqrt{|\cos x(1 - \cos^2 x)|} \, dx = 3 \int_0^{\pi} \sqrt{|\cos x| \sin^2 x} \, dx = 3 \int_0^{\pi} \sin x \sqrt{|\cos x|} \, dx$. Splitting the integral at $x = \frac{\pi}{2}$: $I = 3 \left( \int_0^{\frac{\pi}{2}} \sin x \sqrt{\cos x} \, dx + \int_{\frac{\pi}{2}}^{\pi} \sin x \sqrt{-\cos x} \, dx ight)$. For the first part, let $u = \cos x$, $du = -\sin x \, dx$: $3 \int_0^1 u^{1/2} \, du = 3 \left[ \frac{u^{3/2}}{3/2} ight]_0^1 = 3 	imes \frac{2}{3} = 2$. For the second part, let $u = \cos x$, $du = -\sin x \, dx$: $3 \int_{-1}^0 \sqrt{-u} \, du = 3 \int_0^1 \sqrt{v} \, dv = 3 \left[ \frac{v^{3/2}}{3/2} ight]_0^1 = 2$. Thus, $I = 2 + 2 = 4$.

$\frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$

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