Question

$\int \frac{dx}{1+\sin x+\cos x}$ is equal to

• A. $\log |tan\frac{x}{2}|+c$
• B. $\log |1+tan\frac{x}{2}|+c$
• C. $\log |\cot \frac{x}{2}|+c$
• D. $\log |1+\cot \frac{x}{2}|+c$

Answer 1

The correct option is B $\log |1+tan\frac{x}{2}|+c$

$I=\int \frac{dx}{1+\sin x+\cos x}$
$=\int \frac{1}{1+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}dx$
$=\int \frac{se{c}^2\frac{x}{2}}{2(1+\tan \frac{x}{2})}dx$
Put $1+\tan \frac{x}{2}=t$
$\Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt$
$I=\int \frac{dt}{t}$
$=\log |t|+C$
$I=\log |1+\tan \frac{x}{2}|+C$