UPSC NDA 2 2016 Mathematics Solved Paper

Concept:Use the substitution $t = \tan(x/2)$ to convert the trigonometric integral into a rational form, then integrate using $\int\frac{dt}{a^2 + t^2} = \frac{1}{a}\tan^{-1}\frac{t}{a}$ and finally apply $\tan^{-1}(1/x) = \cot^{-1}x$.Explanation:Let $I = \int_{0}^{\pi/2} \frac{dx}{3\cos x + 5}$.Use the identity $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$ and also $dx = \frac{2 dt}{1+t^2}$ with $t = \tan(x/2)$.After substitution and simplification, the integral becomes $I = \int_{0}^{\pi/2} \frac{\sec^2(x/2)\, dx}{8 + 2\tan^2(x/2)}$.Let $t = \tan(x/2)$; then $\sec^2(x/2)dx = 2\,dt$. The limits change: when $x=0$, $t=0$; when $x=\pi/2$, $t=1$.Thus $I = \frac12 \int_{0}^{1} \frac{2\,dt}{4 + t^2} = \int_{0}^{1} \frac{dt}{2^2 + t^2}$.Now integrate: $I = \left[ \frac12 \tan^{-1}\frac{t}{2} \right]_{0}^{1} = \frac12 \tan^{-1}\frac12$.Since $\tan^{-1}\frac12 = \cot^{-1}2$, we obtain $I = \frac12 \cot^{-1}2$.Comparing with $k \cot^{-1}2$ gives $k = \frac12$.Answer:$k = \frac12$ (Option B).