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$$\int_{0}^{\pi} \frac{x \cos x \cdot \sin x}{\cos ^{3} x+\cos x} d x=$$
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The correct answer is:
$\frac{\pi}{8}$
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 $x=\pi, t=-1$
$2 I=-\int_{1}^{-1} \frac{\pi d t}{1+t^{2}}=\pi \int_{-1}^{1} \frac{d t}{1+t^{2}}=2 \pi \int_{0}^{1} \frac{d t}{1+t^{2}}$
$2 I=2 \pi\left[\tan ^{-1} t\right]_{0}^{1}=2 \pi\left(\frac{\pi}{4}\right)=\left(\frac{\pi^{2}}{2}\right)$
$\therefore \frac{\pi^{2}}{4}$
$\quad=\int_{0}^{\pi} \frac{x \sin x}{\cos ^{2} x+1} d x$
$\quad=\int_{0}^{\pi} \frac{(\pi-x) \sin x}{\cos ^{2} x+1} d x=\int_{0}^{\pi} \frac{\pi \sin x}{\cos ^{2} x+1} d x-\int_{0}^{\pi} \frac{x \sin x}{\cos ^{2} x+1} d x$
$\quad=\int_{0}^{\pi} \frac{\pi \sin x}{\cos ^{2} x+1} d x-I$
$2 I=\int_{0}^{\pi} \frac{\pi \sin x}{\cos ^{2} x+1} d x$
Put $\cos x=t=\sin d x=-d t$
When $x=0, t=1$ and when $x=\pi, t=-1$
$2 I=-\int_{1}^{-1} \frac{\pi d t}{1+t^{2}}=\pi \int_{-1}^{1} \frac{d t}{1+t^{2}}=2 \pi \int_{0}^{1} \frac{d t}{1+t^{2}}$
$2 I=2 \pi\left[\tan ^{-1} t\right]_{0}^{1}=2 \pi\left(\frac{\pi}{4}\right)=\left(\frac{\pi^{2}}{2}\right)$
$\therefore \frac{\pi^{2}}{4}$
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