Question

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

• A. $\tan x-\sec x+C$
• B. $\tan x+\sec x+C$
• C. ${\tan }^{2}x+\sec x+C$
• D. $\tan x+{\sec }^{2}x+C$

Answer 1

The correct option is A $\tan x-\sec x+C$

$\int \frac{1}{1+\sin x}dx$
Dividing and multiplying by $(1-\sin x)$
$\int \left(\frac{1}{1+\sin x}\right)\cdot \frac{(1-\sin x)}{(1-\sin x)}dx$
$\{\because 1-{\sin }^{2}x={\cos }^{2}x\}$
$=\int \frac{1-\sin x}{{\cos }^{2}x}dx$
$=\int \frac{1}{{\cos }^{2}x}dx-\int \frac{\sin x}{{\cos }^{2}x}dx$
$=\int {\sec }^{2}xdx-\int \tan x\sec xdx$
$=\tan x-\sec x+C$