Question

$\int \frac{1}{{\tan }^{2}x+{\sec }^{2}x}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\tan x\right)-{\tan }^{-1}\left(\tan x\right)+C$
• B. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\tan x\right)-{\tan }^{-1}\left(\sqrt{2}\tan x\right)+C$
• C. $\sqrt{2}{\tan }^{-1}\left(\sqrt{2}\tan x\right)-{\tan }^{-1}\left(\tan x\right)+C$
• D. $\sqrt{2}{\tan }^{-1}\left(\sqrt{2}\tan x\right)-{\tan }^{-1}\left(\sqrt{2}\tan x\right)+C$

Answer 1

The correct option is C $\sqrt{2}{\tan }^{-1}\left(\sqrt{2}\tan x\right)-{\tan }^{-1}\left(\tan x\right)+C$

Let $I=\int \frac{1}{{\tan }^{2}x+{\sec }^{2}x}dx$
Multiplying ${\sec }^{2}x$ in numerator and denominator
$\Rightarrow I=\int \frac{{\sec }^{2}x}{({\tan }^{2}x+{\sec }^{2}x){\sec }^{2}x}dx$
$\Rightarrow I=\int \frac{{\sec }^{2}x}{(1+2{\tan }^{2}x)(1+{\tan }^{2}x)}dx$
Put $\sqrt{2}\tan x=t$
$\Rightarrow \sqrt{2}{\sec }^{2}xdx=dt$
$\therefore I=\frac{1}{\sqrt{2}}\int \frac{dt}{(1+t^2)(1+\frac{t^2}{2})}$
$\Rightarrow I=\int \frac{\sqrt{2}dt}{(t^2+1)(t^2+2)}$
$\Rightarrow I=\sqrt{2}\left[\int \frac{dt}{1+t^2}-\int \frac{dt}{2+t^2}\right]$
$\{\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\}$
$\Rightarrow I=\sqrt{2}[{\tan }^{-1}t-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{t}{\sqrt{2}}]+C$
$\therefore I=\sqrt{2}{\tan }^{-1}\left(\sqrt{2}\tan x\right)-{\tan }^{-1}\left(\tan x\right)+C$