JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 6)
The integral $$\int\limits_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} \mathrm{~d} x$$ is equal to :
$$\tan ^{-1}(2)$$
$$\tan ^{-1}(2)-\frac{\pi}{4}$$
$$\frac{1}{2} \tan ^{-1}(2)-\frac{\pi}{8}$$
$$\frac{1}{2}$$
Explanation
$$I = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx} $$
$$ = \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}} $$
Let $$\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt$$
$$I = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}} $$
$$ = \int\limits_0^1 {{{dt} \over {{t^2} + 2t + 2}} = \int\limits_0^1 {{{dt} \over {{{(t + 1)}^2} + 1}}} } $$
$$ = \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}$$
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