Verified Solution • 2025–26 Session moderate QJEEM22Ma516d: JEE Main 2022 Mathematics - Class 12 JEEM22Ma516d Solution: The integral is equal to : The integral ∫0π213+2sinx+cosx dx\int\limits_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} \mathrm{~d} x0∫2π3+2sinx+cosx1 dx is equal to : Multiple Choice Options A tan−1(2)\tan ^{-1}(2)tan−1(2) B tan−1(2)−π4\tan ^{-1}(2)-\frac{\pi}{4}tan−1(2)−4π C 12tan−1(2)−π8\frac{1}{2} \tan ^{-1}(2)-\frac{\pi}{8}21tan−1(2)−8π D 12\frac{1}{2}21 Step-by-Step Verified Solution I=∫0π/213+2sinx+cosxdxI = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx}I=0∫π/23+2sinx+cosx1dx =∫0π/2(1+tan2x/2)dx3(1+tan2x/2)+2(2tanx/2)+(1−tan2x/2)= \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}}=0∫π/23(1+tan2x/2)+2(2tanx/2)+(1−tan2x/2)(1+tan2x/2)dx Let tanx/2=t⇒sec2x/2dx=2dt\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dttanx/2=t⇒sec2x/2dx=2dt I=∫012dt4+2t2+4tI = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}}I=0∫14+2t2+4t2dt =∫01dtt2+2t+2=∫01dt(t+1)2+1= \int\limits_0^1 {{{dt} \over {{t^2} + 2t + 2}} = \int\limits_0^1 {{{dt} \over {{{(t + 1)}^2} + 1}}} }=0∫1t2+2t+2dt=0∫1(t+1)2+1dt =tan−1(t+1)∣01=tan−12−π4= \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}=tan−1(t+1)01=tan−12−4π Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring