Practicing Success
\(\int \frac{dx}{1+\sin x}\) is equal to |
\(\sec x+\tan x+C\) \(\tan x-\sec x+C\) \(\sec x-\tan x+C\) None |
\(\tan x-\sec x+C\) |
$I=\int \frac{1}{1+\sin x}×\frac{1-\sin x}{1-\sin x}dx=\int\frac{1+\sin x}{\cos^2x}dx=\int\sec^2x-\sec x\tan xdx$ $=\tan x-\sec x+C$ |