\(\int\displaystyle\frac{1}{1+\sin x}\,dx=\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\tan x-\sec x+C\)
Multiplying the numerator and the denominator by \(1-\sin x\) (the conjugate):
\( \displaystyle \frac{1}{1+\sin x} = \frac{1-\sin x}{1-\sin^2x} \)
\( \displaystyle = \frac{1-\sin x}{\cos^2x} \)
\( \displaystyle = \sec^2x-\sec x\tan x \)
Integrating with the standard integrals \(\int\sec^2x\,dx=\tan x\) and \(\int\sec x\tan x\,dx=\sec x\):
\( \displaystyle \int\frac{dx}{1+\sin x} = \tan x-\sec x+C \)