Verified Solution • 2025–26 Session moderate

QJEEM22Ma516d: JEE Main 2022 Mathematics - Class 12 JEEM22Ma516d Solution: The integral is equal to :

The integral ∫0π213+2sin⁡x+cos⁡x dx\int\limits_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} \mathrm{~d} x is equal to :

Multiple Choice Options

  • A tan⁡−1(2)\tan ^{-1}(2)
  • B tan⁡−1(2)−π4\tan ^{-1}(2)-\frac{\pi}{4}
  • C 12tan⁡−1(2)−π8\frac{1}{2} \tan ^{-1}(2)-\frac{\pi}{8}
  • D 12\frac{1}{2}

Step-by-Step Verified Solution

I=∫0π/213+2sin⁡x+cos⁡xdxI = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx}

=∫0π/2(1+tan⁡2x/2)dx3(1+tan⁡2x/2)+2(2tan⁡x/2)+(1−tan⁡2x/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)}}}

Let tan⁡x/2=t⇒sec⁡2x/2dx=2dt\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt

I=∫012dt4+2t2+4tI = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}}

=∫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}}} }

=tan⁡−1(t+1)∣01=tan⁡−12−π4= \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}

Related Learning Resources & Exam Preparation