From the given ∫(11+cotx)dx
If an integrand is a rational function of the trigonometric functions, the substitution z=tan(x2), or its equivalent
sinx=2z1+z2 and cosx=1−z21+z2 and
dx=2dz1+z2
The solution:
∫(11+cotx)dx
∫(11+cosxsinx)dx
∫(sinxsinx+cosx)dx
∫2z1+z2(2z1+z2+1−z21+z2)⋅(2dz1+z2)
Simplify
∫2z1+z2(2z1+z2+1−z21+z2)⋅(2dz1+z2)
∫4z(−z2+2z+1)(z2+1)⋅dz
∫−4z(z2+1)(z2−2z−1)⋅dz
At this point, use Partial Fractions then integrate
∫−4z(z2+1)(z2−2z−1)⋅dz=∫(Az+Bz2+1+Cz+Dz2−2z−1)dz
We do the Partial Fractions first
−4z(z2+1)(z2−2z−1)=Az+Bz2+1+Cz+Dz2−2z−1
−4z(z2+1)(z2−2z−1)=(Az+B)(z2−2z−1)+(Cz+D)(z2+1)(z2+1)(z2−2z−1)
Expand the right side of the equation
−4z(z2+1)(z2−2z−1)=
Az3−2Az2−Az+Bz2−2Bz−B+Cz3+Dz2+Cz+D(z2+1)(z2−2z−1)
Set up the equations
0⋅z3+0⋅z2−4⋅z+0⋅z0(z2+1)(z2−2z−1)=
(A+C)⋅z3+(−2A+B+D)⋅z2+(−A−2B+C)⋅z+(−B+D)⋅z0(z2+1)(z2−2z−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
∫−4z(z2+1)(z2−2z−1)⋅dz=∫(Az+Bz2+1+Cz+Dz2−2z−1)dz=∫(z+1z2+1+−z+1z2−2z−1)dz=
12∫2zz2+1dz+∫dzz2+1−12∫2z−2z2−2z−1dz
=12⋅ln(z2+1)+tan−1z−12⋅ln(z2−2z−1)
=12⋅ln(z2+1z2−2z−1)+tan−1z
We will return it to its original variable x using z=tan(x2) for the final answer.
∫(11+cotx)dx=
12⋅ln(tan2(x2)+1tan2(x2)−2⋅tan(x2)−1)+x2+K
where K= constant of integration
God bless...I hope the explanation is useful.
