How to integrate $\int\frac{1}{1+\sin(x)} dx$ $$\int\frac{1}{1+\sin(x)} dx$$
The integration techniques I know are:
Inspection, power rule, integral of basic functions (also trig), and substitution
But none of them help me solve this.
How can I solve this integral with the techniques I know so far?
 A: Hint: Note that
$$\frac{1}{1+\sin x}=\frac{1-\sin x}{1-\sin^2 x}=\frac{1-\sin x}{\cos ^2x}=\sec^2x-\sec x\tan x.$$
Then recall that
$$(\tan x)'=\sec^2x,\quad (\sec x)'=\sec x\tan x.$$
A: HINT:
$$1+\sin x=1+\cos\left(\dfrac\pi2-x\right)=2\cos^2\left(\dfrac\pi4-\dfrac x2\right)$$
A: HINT
Let $t=\tan \frac{x}{2}$. Now using Weierstrass substitution
Note that the integral becomes $$\int \frac{1}{1+\sin x} \mathrm{d}x=\int \frac{2}{t^2+2t+1}  \mathrm {d} t$$
Can you take it from here?
A: $$\displaylines{
  \sin x = {{2\tan x/2} \over {1 + {{\tan }^2}x/2}} \cr 
  1 + {\tan ^2}x/2 = {\sec ^2}x/2 \cr}$$
use these formulas to convert the integral to required form.
A: $$\frac 1{1+\sin x}=\frac{1}{(\sin \frac x2+\cos\frac x2)^2}=\frac{1}{\sin^2\frac x2(1+\cot\frac x2)^2}=\frac{1+\cot^2\frac x2}{(1+\cot\frac x2)^2}$$ 
set $u=1+\cot\frac x2$, thus
$$I=2\int-\frac{1}{u^2}du=\frac2u+c$$
A: Use Identity 
$$\displaylines{
  \sin 2x = {{2\tan x} \over {1 + {{\tan }^2}x}} \cr 
  1 + {\tan ^2}x = {\sec ^2}x \cr} $$
Given integral
$$\displaylines{
 I= \int {{1 \over {1 + \sin x}}dx}  = \int {{1 \over {1 + {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}}}}  \cr 
   = \int {{{1 + {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2} + 2\tan {x \over 2}}}} dx \cr} $$ 
Now take 
$$\displaylines{
  \tan {x \over 2} = t \cr 
  {1 \over 2}{\sec ^2}{x \over 2}dx = dt \cr 
  {\sec ^2}{x \over 2}dx = 2dt \cr} $$
this will modify the integral as, 
 $$\displaylines{
  I = \int {{{2dt} \over {1 + {t^2} + 2t}}}  \cr 
   = 2\int {{{dt} \over {1 + {t^2} + 2t}}}  \cr} $$ 
You can integrate this integral. Work on it. Good Luck!
A: You'll probably want to do it by substitution.
Let the denominator equal to say $u$.
You then have a simple integral of the form $\dfrac1u$ which resolves to $\ln u + c$.
A: \begin{align}
I=\int\frac{dx}{1+\sin x}&=\int d\left(\frac{-\cos x}{1+\sin x}\right)=\frac{-\cos x}{1+\sin x}+C
\end{align}
