Evaluate: `int1/(13+3cosx+4sinx)\ dx`

Explanation

`int1/(13+3cosx+4sinx)\ dx`

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

`=int frac{1+tan^2x/2}{13+13tan^2x/2+8tan x/2+3-3tan^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`

`=1/5 int frac{1}{(t+2/5)^2+(6)^2/5}dt`

`=1/6 tan^(-1)(frac{5 tan x/2+2}{6}+C)`