Question

$\int \frac{1}{2-3\cos 2x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{2-3·\cos 2x}dx \\ =\int \frac{1}{2-3\left(2{\cos }^{2}x-1\right)}dx \\ =\int \frac{1}{2-6{\cos }^{2}x+3}dx \\ =\int \frac{1}{5-6{\cos }^{2}x}dx \end{gathered}$$
Dividing numerator and denominator by cos²x, we get
$$\begin{gathered} I=\int \frac{{\sec }^{2}x}{5{\sec }^{2}x-6}dx \\ =\int \frac{{\sec }^{2}x}{5\left(1+{\tan }^{2}x\right)-6}dx \\ =\int \frac{{\sec }^{2}x}{5{\tan }^{2}x-1}dx \\ \mathrm{Putting}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int \frac{1}{5t^2-1}dt \\ =\frac{1}{5}\int \frac{1}{t^2-{\left({\displaystyle \frac{1}{\sqrt{5}}}\right)}^2}dt \\ =\frac{1}{5}\times \frac{1}{2\times {\displaystyle \frac{1}{\sqrt{5}}}}\ln \left|\frac{t-{\displaystyle \frac{1}{\sqrt{5}}}}{t+{\displaystyle \frac{1}{\sqrt{5}}}}\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \int \frac{1}{x^{\mathit{2}}\mathit{-}{a}^{\mathit{2}}}dx\mathit{=}\frac{1}{2a}\ln \left|\frac{x\mathit{-}a}{x+a}\right|+C\right] \\ =\frac{1}{2\sqrt{5}}\ln \left|\frac{\sqrt{5}t-1}{\sqrt{5}t+1}\right|+C \\ =\frac{1}{2\sqrt{5}}\ln \left|\frac{\sqrt{5}\tan x-1}{\sqrt{5}\tan x+1}\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathit{\because }\mathit{}t=\tan \mathit{}x\right] \end{gathered}$$