Question

`int 1/(3+5sinx+3cosx)dx`?

Answer 1

`I=1/5ln|5tan(x/2)+3|+c`

Explanation:

Here,

`I=int1/(3+5sinx+3cosx)dx`

Subst. `tan(x/2)=t=>sec^2(x/2)*1/2dx=dt`

`=>(1+tan^2(x/2))dx=2dt=>dx=2/(1+t^2)dt`

So,

`I=int1/(3+5((2t)/(1+t^2))+3((1-t^2)/(1+t^2)))xx2/(1+t^2)dt`

`=int2/(3+3t^2+10t++3-3t^2)dt`

`=int2/(10t+6)dt`

`=int1/(5t+3)dt`

`=1/5ln|5t+3|+c`

Subst. back , `t=tan(x/2)`

`I=1/5ln|5tan(x/2)+3|+c`