Question

The value of $\int \frac{dx}{5+4\cos x}$ is
(where $C$ is integration constant)

• A. $\frac{5}{4}{\tan }^{-1}\left(\frac{1}{3}\tan \frac{x}{2}\right)+C$
• B. $\frac{5}{4}{\cot }^{-1}\left(\frac{1}{3}\cot \frac{x}{2}\right)+C$
• C. $\frac{2}{3}{\cot }^{-1}\left(\frac{1}{3}\cot \frac{x}{2}\right)+C$
• D. $\frac{2}{3}{\tan }^{-1}\left(\frac{1}{3}\tan \frac{x}{2}\right)+C$

Answer 1

The correct option is D $\frac{2}{3}{\tan }^{-1}\left(\frac{1}{3}\tan \frac{x}{2}\right)+C$

$I=\int \frac{dx}{5+4\cos x}=\int \frac{dx}{5+4\frac{(1-{\tan }^2\frac{x}{2})}{(1+{\tan }^2\frac{x}{2})}}[\because \cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}]\Rightarrow I=\int \frac{{\sec }^2\frac{x}{2}dx}{9+{\tan }^2\frac{x}{2}}$
Put $\tan \frac{x}{2}=t$
$\Rightarrow {\sec }^2\frac{x}{2}dx=2dt\Rightarrow I=\int \frac{2dt}{9+t^2}=\frac{2}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+C=\frac{2}{3}{\tan }^{-1}\left(\frac{\tan \frac{x}{2}}{3}\right)+C$