Question

$\int \frac{dx}{4+5\cos \frac{x}{2}}$ is equal to
(where $C$ is integration constant)

• A. $\frac{2}{3}\ln \left|\frac{3+\tan \frac{x}{4}}{3-\tan \frac{x}{4}}\right|+C$
• B. $\frac{3}{2}\ln \left|\frac{3-\tan \frac{x}{4}}{3+\tan \frac{x}{4}}\right|+C$
• C. $\frac{2}{3}\ln \left|\frac{3-\tan \frac{x}{4}}{3+\tan \frac{x}{4}}\right|+C$
• D. $\frac{3}{2}\ln \left|\frac{3+\tan \frac{x}{4}}{3-\tan \frac{x}{4}}\right|+C$

Answer 1

The correct option is A $\frac{2}{3}\ln \left|\frac{3+\tan \frac{x}{4}}{3-\tan \frac{x}{4}}\right|+C$

$I=\int \frac{dx}{4+5\cos \frac{x}{2}}$
Put, $t=\tan \frac{x}{4}$
$\Rightarrow x=4{\tan }^{-1}t\Rightarrow dx=\frac{4dt}{1+t^2}\cos \frac{x}{2}=\frac{1-t^2}{1+t^2}\Rightarrow I=\int \frac{\frac{4dt}{1+t^2}}{4+5\cdot \frac{1-t^2}{1+t^2}}=\int \frac{4dt}{4(1+t^2)+5(1-t^2)}=4\int \frac{dt}{3^2-t^2}=4\cdot \frac{1}{2\cdot 3}\ln \left|\frac{3+t}{3-t}\right|+C=\frac{2}{3}\ln \left|\frac{3+\tan \frac{x}{4}}{3-\tan \frac{x}{4}}\right|+C$