Question

Integrate : $\int \frac{\cos 2x}{1+{\sin }^{2}x}dx$

Answer 1

$I=\int \frac{\cos 2x}{1+{\sin }^{2}x}dx=\int \frac{\cos 2x}{1+\frac{1-\cos 2x}{2}}dx$ [Using $\cos 2x=1-2{\sin }^{2}x]$

$I=-2\int \frac{-\cos 2x}{3-\cos 2x}$ [Adding and subtracting both sides]

$I=-2\int \frac{(3-\cos 2x)-3}{3-\cos 2x}dx$ [Splitting the fraction]

$I=-2\int 1.dx+6\int \frac{1}{3-\cos 2x}dx$

$I=-2x+6\int \frac{1}{3-\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}}dx[Usingrelation\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}]$

$I=-2x+6\int \frac{{\sec }^{2}x}{2+4{\tan }^{2}x}dx[{\sec }^{2}x=1+{\tan }^{2}x]$

let $\tan x=t$

Differentiation both sides:-

${\sec }^{2}xdx=dt$

Subtracting in the integral:-

$\Rightarrow I=-2x+6^3\int \frac{dt}{x(1+2t^2)}$

$=-2x+3\int \frac{dt}{1+\left(\sqrt{2}{t}^2\right)}[integraloffrom\frac{1}{1+x^2}\because \int \frac{1}{1+x^2}dx={\tan }^{-1}x+c]$

$=-2x+\frac{3}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}t\right)+c$

$\Rightarrow I=-2x+\frac{3}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\tan x\right)+c$