From the given int (1/(1+cot x)) dx
If an integrand is a rational function of the trigonometric functions, the substitution z=tan (x/2), or its equivalent
sin x=(2z)/(1+z^2) and cos x=(1-z^2)/(1+z^2) and
dx=(2dz)/(1+z^2)
The solution:
int (1/(1+cot x)) dx
int (1/(1+cos x/sin x)) dx
int (sin x/(sin x+cos x)) dx
int ((2z)/(1+z^2))/(((2z)/(1+z^2)+(1-z^2)/(1+z^2))) *((2dz)/(1+z^2))
Simplify
int ((2z)/(1+z^2))/(((2z)/(1+z^2)+(1-z^2)/(1+z^2))) *((2dz)/(1+z^2))
int (4z)/((-z^2+2z+1)(z^2+1))*dz
int (-4z)/((z^2+1)(z^2-2z-1))*dz
At this point, use Partial Fractions then integrate
int (-4z)/((z^2+1)(z^2-2z-1))*dz=int ((Az+B)/(z^2+1)+(Cz+D)/(z^2-2z-1))dz
We do the Partial Fractions first
(-4z)/((z^2+1)(z^2-2z-1))= (Az+B)/(z^2+1)+(Cz+D)/(z^2-2z-1)
(-4z)/((z^2+1)(z^2-2z-1))= ((Az+B)(z^2-2z-1)+(Cz+D)(z^2+1))/((z^2+1)(z^2-2z-1))
Expand the right side of the equation
(-4z)/((z^2+1)(z^2-2z-1))=
(Az^3-2Az^2-Az+Bz^2-2Bz-B+Cz^3+Dz^2+Cz+D)/((z^2+1)(z^2-2z-1))
Set up the equations
(0*z^3+0*z^2-4*z+0*z^0)/((z^2+1)(z^2-2z-1))=
((A+C)*z^3+(-2A+B+D)*z^2+(-A-2B+C)*z+(-B+D)*z^0)/((z^2+1)(z^2-2z-1))
The equations are
A+C=0
-2A+B+D=0
-A-2B+C=-4
-B+D=0
Simultaneous solution results to
A=1 and B=1 and C=-1 and D=1
We can now do the integration
int (-4z)/((z^2+1)(z^2-2z-1))*dz=int ((Az+B)/(z^2+1)+(Cz+D)/(z^2-2z-1))dz=int ((z+1)/(z^2+1)+(-z+1)/(z^2-2z-1))dz=
1/2 int (2z)/(z^2+1) dz+int dz/(z^2+1)-1/2int (2z-2)/(z^2-2z-1)dz
=1/2*ln (z^2+1)+tan^-1 z-1/2*ln(z^2-2z-1)
=1/2*ln((z^2+1)/(z^2-2z-1))+tan^-1 z
We will return it to its original variable x using z=tan (x/2) for the final answer.
color(blue)(int (1/(1+cot x)) dx=)
color(blue)(1/2*ln((tan^2 (x/2)+1)/(tan^2 (x/2)-2*tan (x/2)-1))+x/2+K)
where K= constant of integration
God bless...I hope the explanation is useful.
