(D) Let $I=\int_0^{\frac{\pi}{2}} \frac{d \theta}{1+\tan \theta}$ …(i) $=\int_0^{\frac{\pi}{2}} \frac{d \theta}{1+\tan \left(\frac{\pi}{2}-\theta\right)}$ $\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$ $\therefore \quad I=\int_0^{\frac{\pi}{2}} \frac{d \theta}{1+\cot \theta}$ ...(ii) Adding (i) and (ii), we get $2 I=\int_0^{\frac{\pi}{2}}\left(\frac{1}{1+\tan \theta}+\frac{1}{1+\cot \theta}\right) d \theta$ $=\int_0^{\frac{\pi}{2}}\left(\frac{1}{1+\tan \theta}+\frac{\tan \theta}{\tan \theta+1}\right) d \theta$ $=\int_0^{\frac{\pi}{2}} d \theta=[\theta]_0^{\pi / 2}$ $\therefore \quad 2 I =\frac{\pi}{2} \Rightarrow I =\frac{\pi}{4}$ Alternate Method: $\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan ^{ n } x}=\int_0^{\frac{\pi}{2}} \frac{d x}{1+\cot ^{ n } x} d x=\frac{\pi}{4}$ $\int_0^{\pi / 2} \frac{d \theta}{1+\tan \theta}=\frac{\pi}{4}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.