Question

Solve the integralI've come across some difficulties with this one..

Answer 1

Multiply the integrand's numerator and denominator by $1-\sin x$, then simplify:

$\displaystyle\int\frac{\mathrm dx}{1+\sin x}=\int\frac{1-\sin x}{1-\sin^2x}\,\mathrm dx=\int\frac{1-\sin x}{\cos^2x}\,\mathrm dx=\int(\sec^2x-\sec x\tan x)\,\mathrm dx$

You should be familiar with

$\dfrac{\mathrm d}{\mathrm dx}\tan x=\sec^2x$

$\dfrac{\mathrm d}{\mathrm dx}\sec x=\sec x\tan x$

so that

$\displaystyle\int\frac{\mathrm dx}{1+\sin x}=\tan x-\sec x+C$