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

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