Question

Evaluate: $\int \frac{dx}{1-\sin x}$

• A. $x+\cos x+c$
• B. $1+\sin x+c$
• C. $\sec x-\tan x+c$
• D. $\sec x+\tan x+c$

Answer 1

The correct option is D $\sec x+\tan x+c$

$I=\int \frac{dx}{1-\sin x}$

$=\int \frac{dx}{1-\sin x}\frac{1+\sin x}{1+\sin x}$

$=\int \frac{1+\sin x}{1-{\sin }^{2}x}dx$

$=\int \frac{1+\sin x}{{\cos }^{2}x}dx$

$=\int \frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x}dx$

$=\int {\sec }^{2}x+\sec x\tan xdx$

$=\tan x+\sec x+c$