Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

If \( A = \int_{0}^{\pi} \frac{cos x}{(x+2)^2}dx\), then show that \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(x+1)}dx = \frac{1}{2} \big( \frac{1}{2} +\frac{1}{\pi+2} - A\big)\). $$$$ Solution: \( A = \cos x \int_{0}^{\pi} \frac{dx}{(x+2)^2} - \int_{0}^{\pi} \bigg( \frac{d}{dx} \cos x \int \frac{dx}{(x+2)^2} \bigg) dx \) $$$$ \( \implies A = \frac{-\cos x}{x+2}\mid_{0}^{\pi} - \int_{0}^{\pi} \frac{\sin x}{(x+2)} dx \) $$$$ \( \implies A = \frac{1}{\pi +2} + \frac{1}{2} - \int_{0}^{\pi} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2} }{2(1+\frac{x}{2})} dx \) $$$$ Putting $\frac{x}{2}=t$ in the last integral, we have $dx=2dt$ and $t$ varies from $0$ to $\frac{\pi}{2}$ $$$$ Therefore, \( A = \frac{1}{\pi +2} + \frac{1}{2} - 2 \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt \)$$$$ \( \implies \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$ \( i.e., \) \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(1+x)}dx = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$