Question

The value of $\int \frac{dx}{1+3{\cos }^{2}x}$
(where $C$ is integration constant)

• A. ${\tan }^{-1}\left(\frac{\tan x}{2}\right)+C$
• B. $\frac{1}{2}{\tan }^{-1}\left(\frac{\tan x}{2}\right)+C$
• C. $\frac{1}{3}{\tan }^{-1}\left(\frac{\tan x}{3}\right)+C$
• D. ${\tan }^{-1}\left(\frac{\tan x}{3}\right)+C$

Answer 1

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

Multiplying Numerator and Denominator of the given integral by ${\sec }^{2}x$, we get
$I=\int \frac{{\sec }^{2}xdx}{{\sec }^{2}x+3}\Rightarrow I=\int \frac{{\sec }^{2}xdx}{{\tan }^{2}x+4}$
Put $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt\Rightarrow I=\int \frac{dt}{t^2+2^2}\Rightarrow I=\frac{1}{2}{\tan }^{-1}\frac{t}{2}+C[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C]\therefore I=\frac{1}{2}{\tan }^{-1}\left(\frac{\tan x}{2}\right)+C$