\(\int \frac{dx}{1+\sin x}\) is equal to
Answer & explanation
Correct answer: option 2
$I=\int \frac{1}{1+\sin x}×\frac{1-\sin x}{1-\sin x}dx=\int\frac{1+\sin x}{\cos^2x}dx=\int\sec^2x-\sec x\tan xdx$
$=\tan x-\sec x+C$
\(\int \frac{dx}{1+\sin x}\) is equal to
Correct answer: option 2
$I=\int \frac{1}{1+\sin x}×\frac{1-\sin x}{1-\sin x}dx=\int\frac{1+\sin x}{\cos^2x}dx=\int\sec^2x-\sec x\tan xdx$
$=\tan x-\sec x+C$