int_{0}^{pi} frac{x cos x sin x}{cos^{3} x + | vedclass

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(B) Let $I = \int_{0}^{\pi} \frac{x \cos x \sin x}{\cos^{3} x + \cos x} dx$. Simplifying the integrand: $I = \int_{0}^{\pi} \frac{x \sin x}{\cos^{2} x

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$\frac{\pi^{2}}{4}$

Vedclass · Answer

(B) Let $I = \int_{0}^{\pi} \frac{x \cos x \sin x}{\cos^{3} x + \cos x} dx$. Simplifying the integrand: $I = \int_{0}^{\pi} \frac{x \sin x}{\cos^{2} x + 1} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$: $I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{\cos^{2}(\pi - x) + 1} dx = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{\cos^{2} x + 1} dx$. $I = \pi \int_{0}^{\pi} \frac{\sin x}{\cos^{2} x + 1} dx - \int_{0}^{\pi} \frac{x \sin x}{\cos^{2} x + 1} dx$. $I = \pi \int_{0}^{\pi} \frac{\sin x}{\cos^{2} x + 1} dx - I$. $2I = \pi \int_{0}^{\pi} \frac{\sin x}{\cos^{2} x + 1} dx$. Let $t = \cos x$,then $dt = -\sin x dx$. When $x = 0, t = 1$; when $x = \pi, t = -1$. $2I = -\pi \int_{1}^{-1} \frac{dt}{t^{2} + 1} = \pi \int_{-1}^{1} \frac{dt}{t^{2} + 1} = 2\pi \int_{0}^{1} \frac{dt}{t^{2} + 1}$. $2I = 2\pi [	an^{-1} t]_{0}^{1} = 2\pi (\frac{\pi}{4} - 0) = \frac{\pi^{2}}{2}$. $I = \frac{\pi^{2}}{4}$.

$\int_{0}^{\pi} \frac{x \cos x \sin x}{\cos^{3} x + \cos x} dx = $

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