$\int \frac{1}{3-2 \cos 2 x} \mathrm{~d} x=$ (where $C$ is constant of integration.)

$\int \frac{1}{3-2 \cos 2 x} \mathrm{~d} x=$ (where $C$ is constant of integration.)
  1. $\frac{2}{5} \tan ^{-1}(5 \tan x)+C$
  2. $\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \tan x)+C$
  3. $\frac{2}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \tan x)+C$
  4. $\frac{1}{5} \tan ^{-1}(5 \tan x)+C$

Solution

$\begin{aligned} & \int \frac{\mathrm{d} x}{3-2 \cos 2 x}=\int \frac{\mathrm{d} x}{3-2 \times \frac{1-\tan ^2 x}{1+\tan ^2 x}}=\int \frac{\sec ^2 x \mathrm{~d} x}{5 \tan ^2 x+1} \\ & =\int \frac{\mathrm{d} t}{5 t^2+1}[\text { let } \tan x=t] \\ & =\int \frac{\mathrm{d} t}{(\sqrt{5} t)^2+1^2}=\frac{1}{\sqrt{5}} \tan ^{-1}\left(\frac{\sqrt{5} t}{1}\right)+C \\ & =\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \tan x)+C\end{aligned}$

Asked in: MHT CET 2022 (10 Aug Shift 2)

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