Question

How do you integrate `1/(1+tanx) dx`?

Answer 1

Use the substitution `tanx=u`.

Explanation:

Let

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

Apply the substitution `tanx=u`:

`I=int1/((1+u^2)(1+u))du`

Apply partial fraction decomposition:

`I=1/2int((1-u)/(1+u^2)+1/(1+u))du`

Rearrange:

`I=1/2int(1/(1+u^2)-1/2(2u)/(1+u^2)+1/(1+u))du`

Integrate term by term:

`I=1/2(tan^(-1)u-1/2ln(1+u^2)+ln(1+u))+C`

Reverse the substitution:

`I=1/2(x-ln(secx)+ln(1+tanx))+C`

Simplify:

`I=1/2(x+ln(sinx+cosx))+C`

Answer 2

`int1/(1+tanx)"d"x=1/4(2x+ln(1+sin2x))+"c"`

Explanation:

We want to find `int1/(1+tanx)"d"x`.

We start by substituting `u=tanx` and `"d"u=sec^2x"d"x=(1+tan^2x)"d"x=(1+u^2)"d"x`

So

`int1/(1+tanx)"d"x=int1/((1+u)(1+u^2))"d"u`

Now we perform a partial fraction decomposition on the integrand

`1/((1+u)(1+u^2))=A/(1+u)+(Bu+C)/(1+u^2)`

`1=A(1+u^2)+(Bu+C)(1+u)`

Letting `u=-1` gives

`1=A(1+(-1)^2)=2ArArrA=1/2`

Letting `u=0` gives

`1=1/2(1+0)+CrArrC=1/2`

Letting `u=1` gives

`1=1/2(2)+(B+1/2)(2)=2B+2rArrB=-1/2`

So,

`1/((1+u)(1+u^2))=1/2(1/(1+u)+(1-u)/(1+u^2))`

and

`int1/((1+u)(1+u^2))"d"x=`

`1/2(int1/(1+u)"d"u+int1/(1+u^2)"d"u-intu/(1+u^2)"d"u)=`

`1/2(lnabs(1+u)+arctanu-1/2ln(1+u^2))+"c"`

Reversing for `u=tanx` gives

`1/4(2lnabs(1+tanx)+2arctan(tanx)+ln(1/(1+tan^2x)))+"c"`

`=1/4(2x+ln(cos^2x+2sinxcosx+sin^2x))+"c"`

`=1/4(2x+ln(1+sin2x))+"c"`

Answer 3

`1/2{x+ln|(cosx+sinx)|}+C`.

Explanation:

Let, `I=int1/(1+tanx)dx=intcosx/(cosx+sinx)dx`.

`:. I=1/2int(2cosx)/(cosx+sinx)dx`,

`=1/2int{(cosx+sinx)+(cosx-sinx)}/(cosx+sinx)dx`,

`=1/2int{(cosx+sinx)/(cosx+sinx)+(cosx-sinx)/(cosx+sinx)}dx`,

`=1/2int{1+(d/dx(cosx+sinx))/(cosx+sinx)dx}`,

`rArr I=1/2{x+ln|(cosx+sinx)|}+C`.