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

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Aicharya · Accepted Answer

Let I $=\int_{0}^{\pi} \frac{x \cos x \sin x}{\cos ^{3} x+\cos x} d x$ $\quad=\int_{0}^{\pi} \frac{x \sin x}{\cos ^{2} x+1} d x$ $\quad=\int_{0}^{\pi} \frac{(\pi-x) \sin x}{\cos ^{2} x+1} d x=\int_{0}^{\pi} \frac{\pi \sin x}{\cos ^{2} x+1} d x-\int_{0}^{\pi} \frac{x \sin x}{\cos ^{2} x+1} d x$ $\quad=\int_{0}^{\pi} \frac{\pi \sin x}{\cos ^{2} x+1} d x-I$ $2 I=\int_{0}^{\pi} \frac{\pi \sin x}{\cos ^{2} x+1} d x$ Put $\cos x=t=\sin d x=-d t$ When $x=0, t=1$ and when $x=\pi, t=-1$ $2 I=-\int_{1}^{-1} \frac{\pi d t}{1+t^{2}}=\pi \int_{-1}^{1} \frac{d t}{1+t^{2}}=2 \pi \int_{0}^{1} \frac{d t}{1+t^{2}}$ $2 I=2 \pi\left[	an ^{-1} tight]_{0}^{1}=2 \pi\left(\frac{\pi}{4}ight)=\left(\frac{\pi^{2}}{2}ight)$ $	herefore \frac{\pi^{2}}{4}$

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

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