Evaluate: `int1/(5+7cosx+sinx)\ dx`

Explanation

`int1/(5+7cosx+sinx)\ dx`

`sinx=frac{2 tan x/2}{1+tan^2 x/2}` and `cosx=frac{1-tan^2 x/2}{1+tan^2 x/2}`

`=int frac{1+tan^2 x/2}{5+5tan^2x/2+2tan x/2+7-7tan^2x/2}dx`

Replacing `1 + tan^2x/2` in numerator by `sec^2x/2` and putting `tan x/2 = t` and `sec^2 x/2 dx = 2dt`

`=-int frac{1}{t^2-t-6}dt`

`=-int frac{1}{(t-1/2)^2-(5)^2/2}dt`

`=-1/5 log abs(frac{tan x/2-3}{tan x/2+2})+C`