Question

How do you find the integral of `int 1/(1 + tan(x))`?

Answer 1

`intcosx/(cos+sinx)dx=1/2(x+ln(cosx+sinx))+c`

Explanation:

Let

`I=int1/(1+tanx)dx`

`I=intcosx/(cos+sinx)dx`

`cosx=l(cosx+sinx)+m(-sinx+cosx)+n`

`cosx=(l+m)cosx+(l-m)sinx+n`

Equating the coefficients of cosx sinx and constants

`l+m=1, l-m=0, n=0`

`l=1/2, n=1/2`

`cosx=1/2(cosx+sinx)+1/2(-sinx+cosx)+0`

`I=intcosx/(cos+sinx)dx`

`I=1/2int(cosx+sinx)/(cosx+sinx)dx+1/2int(-sinx+cosx)/(cosx+sinx)dx`

`I=1/2(I_1+I_2)`

where,

`I_1=int(cosx+sinx)/(cosx+sinx)dx=int1dx=x`

`I_1=x`

`I_2=int(-sinx+cosx)/(cosx+sinx)dx`

let

`t=cosx+sinx`

`dt/dx=-sinx+cosx`

`dt=(-sinx+cosx)dx`

`int(-sinx+cosx)/(cosx+sinx)dx=int((-sinx+cosx)dx)/(cosx+sinx)=intdt/t=lnt`

`I_2=ln(cosx+sinx)`

`I=1/2(I_1+I_2)`

`intcosx/(cos+sinx)dx=1/2(x+ln(cosx+sinx))+c`