Question

$\int \left(\frac{1}{1+cotx}\right)dx$

Answer 1

$\int \left(\frac{1}{1+cotx}\right)dx$
$\int \left(\frac{1}{1+\frac{cosx}{sinx}}\right)dx\int \frac{\frac{2z}{1+z^2}}{(\frac{2z}{1+z^2}+\frac{1-z^2}{1+z^2}).}\left(\frac{2dz}{1+z^2}\right)$
Simplify
$\int \frac{\frac{2z}{1+z^2}}{(\frac{2z}{1+z^2}+\frac{1-z^2}{1+z^2}).}\left(\frac{2dz}{1+z^2}\right)\int \frac{4z}{(z^2+2z+1)(z^2+1).}dz\int \frac{-4z}{(z^2+1)(z^2+2z-1).}dz$
At this point, use Partial Fractions then integrate
$\int \frac{-4z}{(z^2+1)(z^2-2z-1).}dz=\int (\frac{Az+B}{z^2+1}+\frac{Cz+D}{z^2-2z-1})dz$
We do the Partial Fractions first
$\frac{-4z}{(z^2+1)(z^2-2z-1)}=\frac{Az+B}{z^2+1}+\frac{Cz+D}{z^2-2z-1}\frac{-4z}{(z^2+1)(z^2-2z-1)}=\frac{(Az+B)(z^2-2z-1)+(Cz+D)(Cz+D)(z^2+1)}{(z^2+1)(z^2-2z-1)}$
Expand the right side of the equation
$\frac{-4z}{(z^2+1)(z^2-2z-1)}=\frac{A{z}^3-2A{z}^2-Az+B{z}^2-2Bz-B+C{z}^3+D{z}^2+Cz+D}{(z^2+1)(z^2-)(z^2-2z-1)}$
Set up the equations
$\frac{0.z^3+0.z^2-4.z+0.z^{\circ }}{(z^2+1)(z^2-2z-1)}$
$\frac{(A+C).z^3+(-2A+B+D).z^2+(-A-2B+C).z+(-B+D).z^0)}{(z^2+1)(z^2-2z-1)}$
The equations are
$A+C=0-2A+B+D=0-A-2B+C=-4-B+D=0$
Simultaneous solution result to
A = 1 and B = 1 and C = - 1 and D = 1
We can now do the integration
$\int \frac{-4z}{(z^2+1)(z^2-2z-1).}dz=\int (\frac{Az+B}{z^2+1}+\frac{Cz+D}{z^2-2z-1})dz=\int (\frac{z+1}{z^2+1}+\frac{-z+1}{z^2-2z-1})dz=$
$\frac{1}{2}\int \frac{2z}{z^2+1}dz+\int \frac{dz}{z^2+1}-\frac{1}{2}\int \frac{2z-2}{z^2-2z-1}dz=\frac{1}{2}.In(z^2+1)+ta{n}^{-1}z-\frac{1}{2}.In(z^2-2z-1)=\frac{1}{2}.In\left(\frac{z^2+1}{z^2-2z-1}\right)+ta{n}^{-1}z$