Question

How do you integrate dx/(sin2x+tan2x)?

Answer 1

The answer is `=+1/8ln(|1-cos(2x)|)+1/8ln(|1+cos(2x)|)-1/(4(1+cos(2x)))+C`

Explanation:

Perform the substitution

`u=2x`,`=>`, `du=2dx`

Therefore,

`int(dx)/(sin2x+tan2x)=1/2int(du)/(sinu+tanu)`

`tanu=sinu/cosu`

`sin^2u=1-cos^2u`

So,

`int(dx)/(sin2x+tan2x)=1/2int(du)/(sinu+sinu/cosu)`

`=1/2int(cosudu)/(sinucosu+sinu)`

`=1/2int(cosusinudu)/(sin^2u(1+cosu))`

`=1/2int(cosusinudu)/((1-cos^2u)(1+cosu))`

Let `v=cosu`, `=>`, `dv=-sinudu`

Therefore,

`int(dx)/(sin2x+tan2x)=-1/2int(vdv)/((1-v^2)(1+v))`

Perform the partial fraction decomposition

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

`=A/(1-v)+B/(1+v)+C/(1+v)^2`

`=(A(1+v)^2+B(1+v)(1-v)+C(1-v))/((1-v)(1+v)^2)`

Comparig the numerators

`v=A(1+v)^2+B(1+v)(1-v)+C(1-v)`

Let `v=1`, `=>`, `1=4A`, `=>`, `A=1/4`

Let `v=-1`, `=>`, `-1=-2C`, `=>`, `C=1/2`

Coefficients of `v^2`

`0=A-B`, `=>`, `B=-1/4`

Therefore,

`(v)/((1-v^2)(1+v))=(1/4)/(1-v)-(1/4)/(1+v)+(1/2)/(1+v)^2`

So,

`int(dx)/(sin2x+tan2x)=-1/2(int(1/4dv)/(1-v)-int(1/4dv)/(1+v)+int(1/2dv)/(1+v)^2)`

`=+1/8ln(1-v)+1/8ln(1+v)-1/4*1/(1+v)`

`=+1/8ln(|1-cos(2x)|)+1/8ln(|1+cos(2x)|)-1/(4(1+cos(2x)))+C`

Answer 2

`1/4Ln(tanx)-1/8(secx)^2+C`

Explanation:

`int dx/(sin2x+tan2x)`

=`1/2int (2dx)/(sin2x+tan2x)`

After using `y=2x` and `dy=2dx` transforms, this integral became

`1/2int dy/(siny+tany)`

=`1/2int dy/(siny+siny/cosy)`

=`1/2int (cosy*dy)/(siny*cosy+siny)`

=`1/2int (cosy*dy)/(siny*(1+cosy))`

=`1/2int (cosy*siny*dy)/[(siny)^2*(1+cosy)]`

=`1/2int (cosy*siny*dy)/([1-(cosy)^2]*(1+cosy))`

=`1/2int (cosy*siny*dy)/[(1-cosy)*(1+cosy)^2]`

After using `z=cosy` and `dz=-siny*dy` transforms, it became

`-1/2int (z*dz)/[(1-z)*(1+z)^2]`

Now, I decomposed integrand into basic fractions,

`z/[(1-z)*(1+z)^2]=A/(1-z)+B/(1+z)+C/(1+z)^2`

After expanding denominator,

`A*(1+z)^2+B*(1-z^2)+C*(1-z)=z`

Set `z=-1`, `2C=-1`, so `C=-1/2`

Set `z=1`, `4A=1`, so `A=1/4`

Set `z=0`, `A+B+C=0`, so `B=1/4`

Thus,

`-1/2int (z*dz)/[(1-z)*(1+z)^2]`

=`-1/2*[1/4int (dz)/(1-z)+1/4int (dz)/(1+z)-1/2int (dz)/(1+z)^2]`

=`-1/8int (dz)/(1-z)-1/8int (dz)/(1+z)+1/4int (dz)/(1+z)^2`

=`1/8Ln(1-z)-1/8Ln(1+z)-1/4*(1+z)^(-1)+C`

=`1/8Ln((1-z)/(1+z))-1/(4z+4)+C`

=`1/8Ln((1-cosy)/(1+cosy))-1/(4cosy+4)+C`

=`1/8Ln((1-cos2x)/(1+cos2x))-1/(4cos2x+4)+C`

=`1/8Ln[(2(sinx)^2)/(2(cosx)^2)]-1/[8(cosx)^2]+C`

=`1/8Ln[(tanx)^2]-1/8(secx)^2+C`

=`1/4Ln(tanx)-1/8(secx)^2+C`