WOW Saw this integral on Internet and don't know how to do it.

Take the integral:
 integral1/(sin(x) + cos(x) + 1) dx
For the integrand 1/(sin(x) + cos(x) + 1), substitute u = tan(x/2) and du = 1/2 dx sec^2(x/2). Then transform the integrand using the substitutions sin(x) = (2 u)/(u^2 + 1), cos(x) = (1 - u^2)/(u^2 + 1) and dx = (2 du)/(u^2 + 1):
 = integral2/((u^2 + 1) ((2 u)/(u^2 + 1) + (1 - u^2)/(u^2 + 1) + 1)) du
Simplify the integrand 2/((u^2 + 1) ((2 u)/(u^2 + 1) + (1 - u^2)/(u^2 + 1) + 1)) to get 1/(u + 1):
 = integral1/(u + 1) du
For the integrand 1/(u + 1), substitute s = u + 1 and ds = du:
 = integral1/s ds
The integral of 1/s is log(s):
 = log(s) + constant
Substitute back for s = u + 1:
 = log(u + 1) + constant
Substitute back for u = tan(x/2):
 = log(tan(x/2) + 1) + constant
Which is equivalent for restricted x values to:
Answer: |= log(sin(x/2) + cos(x/2)) - log(cos(x/2)) + constant   P.S. log=ln

Sorry Max: I can give you the answer but don't know how to use LaTex!!.

If t = tan(x/2), then tan(x) = 2t/(1 - t^2) and, from the rt-angled triangle, sin(x) = 2t/(1 + t^2) and cos(x) = (1 - t^2)/(1 + t^2) .

dt = (1/2)sec^2(x/2)dx, so dx = 2/(1 + t^2).

On substitution, the (1 + t^2) terms tend to cancel.

Sorry, missed a dt at the end of the last equation.

If t = tan(x/2), then tan(x) = 2t/(1 - t^2) and, from the rt-angled triangle, sin(x) = 2t/(1 + t^2) and cos(x) = (1 - t^2)/(1 + t^2) .

dt = (1/2)sec^2(x/2)dx, so dx = 2/(1 + t^2)dt.

On substitution, the (1 + t^2) terms tend to cancel.

I know that this has already been answered perfectly well, and I have learned from all those answers,

but I want to put up my version  :)

Integrate:

$\displaystyle\int \dfrac{1}{1+\sin x + \cos x}\text{dx}$

Please use LaTeX or Image to reply. I am not used to something like int(1/(1 + sin x + cos x ))dx 

Express in terms of t  where   $t=tan(\frac{x}{2})$

We are suppose to remember what sin(x) and cos(x) equals in terms of t but I never can so I'll derive it.

$\begin{array}{rl} sin(x)&=2sin(\frac{x}{2})cos(\frac{x}{2}) \qquad \qquad cos(x)&=cos^2(\frac{x}{2})-sin^2(\frac{x}{2})\\ &=2tan(\frac{x}{2})cos^2(\frac{x}{2}) \qquad \qquad &=cos^2(\frac{x}{2})[1-tan^2(\frac{x}{2})]\\ &=\frac{2tan(\frac{x}{2})}{sec^2(\frac{x}{2})} \qquad \qquad &=\frac{[1-tan^2(\frac{x}{2})]}{sec^2(\frac{x}{2})} \\ &=\frac{2tan(\frac{x}{2})}{1+tan^2(\frac{x}{2})} \qquad \qquad &=\frac{1-tan^2(\frac{x}{2})}{1+tan^2(\frac{x}{2})} \\ &=\frac{2t}{1+t^2} \qquad \qquad &=\frac{1-t^2}{1+t^2} \\~\\ \end{array}$

so

$\begin{align} 1+sin(x)+cos(x)&=\frac{1+t^2+2t+1-t^2}{1+t^2}\\ &=\frac{1+t^2+2t+1-t^2}{1+t^2}\\ &=\frac{2(t+1)}{1+t^2}\\~\\ \frac{1}{1+sin(x)+cos(x)}&=\frac{1+t^2}{2(t+1)}\\ \end{align}$

$\begin{align} t&=tan(\frac{x}{2})\\ \frac{dt}{dx}&=\frac{1}{2}sec^2(\frac{x}{2})\\ &=\frac{1}{2}\left[1-tan^2(\frac{x}{2})\right]\\ &=\frac{1-t^2}{2}\\ \frac{dx}{dt}&=\frac{2}{1-t^2}\\ dx&=\frac{2}{1-t^2}\;dt \end{align}$

$\begin{align} \displaystyle\int \frac{1}{1+sin(x)+cos(x)}\;dx &=\displaystyle\int \frac{1+t^2}{2(t+1)} \cdot \frac{2}{1+t^2}\;dt\\ &=\displaystyle\int \frac{1}{t+1}\;dt\\~\\ &= ln(1+t)\\ &=ln[1+tan( \frac{x}{2})]+c \end{align}$

Restrictions on the domain.

$\frac{x}{2} \ne n\pi \qquad n\in Z\\ so\\ x\ne 2\pi n\\ also\\ 1+tan(\frac{x}{2})>0\\ \frac{-\pi}{2}+2\pi n<x<\pi+2\pi n \qquad n\in Z$

also

$\quad ln[1+tan( \frac{x}{2})]+c\\ = ln[\frac{cos( \frac{x}{2})+sin( \frac{x}{2})}{cos( \frac{x}{2})}]+c\\ =ln[cos( \frac{x}{2})+sin( \frac{x}{2})]-ln[cos( \frac{x}{2})]+c$

Here is the graph of the solution is you want to play with it.

https://www.desmos.com/calculator/6sbycp3ee2

and here it is just to look at :)    (this is for c=1)

I have included two of the asymptotes.     $x=\pi/2\qquad and \qquad x=\pi$