Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 40

int_(pi/6)^(pi/3)csc^3(x)dx
Apply the integral substitution,
Let u=tan(x/2)
=>du=1/2sec^2(x/2)dx
using pythagorean identity: 1+tan^2(theta)=sec^2(theta)
du=1/2(1+tan^2(x/2))dx
du=1/2(1+u^2)dx
=>dx=(2/(1+u^2))du
csc(x)=1/sin(x)=1/(2sin(x/2)cos(x/2))
=(1/(cos^2(x/2)))/((2sin(x/2)cos(x/2))/(cos^2(x/2)))
=(sec^2(x/2))/(2tan(x/2))
=(1+tan^2(x/2))/(2tan(x/2))
=(1+u^2)/(2u)
Now let's evaluate the indefinite integral,
intcsc^3(x)dx=int((1+u^2)/(2u))^3(2/(1+u^2))du
=int(1+u^2)^2/(4u^3)du
=int(1+2u^2+u^4)/(4u^3)du
=1/4int(1/u^3+2/u+u)du
=1/4(int1/u^3du+2int1/udu+intudu)
=1/4((u^(-3+1)/(-3+1))+2ln|u|+u^2/2)
=1/4(-1/(2u^2)+2ln|u|+u^2/2)
Substitute back u=tan(x/2)
=1/4(-1/(2tan^2(x/2))+2ln|tan(x/2)|+1/2tan^2(x/2)
=1/4(1/2tan^2(x/2)-1/2cot^2(x/2)+2ln|tan(x/2)|)
Add a constant C to the solution,
=1/4(1/2tan^2(x/2)-1/2cot^2(x/2)+2ln|tan(x/2)|)+C
int_(pi/6)^(pi/3)csc^3(x)dx=[1/4(1/2tan^2(x/2)-1/2cot^2(x/2)+2ln|tan(x/2)|)]_(pi/6)^(pi/3)
=[1/4(1/2tan^2(pi/6)-1/2cot^2(pi/6)+2ln|tan(pi/6)|]-[1/4(1/2tan^2(pi/12)-1/2cot^2(pi/12)+2ln|tan(pi/12)|]
=[1/4(1/2(1/sqrt(3))^2-1/2(sqrt(3))^2+2ln(1/sqrt(3))]-[1/4(1/2(2-sqrt(3))^2-1/2(2+sqrt(3))^2+2ln(2-sqrt(3))]
=[1/4(1/6-3/2-ln3)]-[1/4(1/2(4-4sqrt(3)+3)-1/2(4+4sqrt(3)+3+2ln(2-sqrt(3)))]
=[-1/3-ln(3)/4]-[1/4(7/2-2sqrt(3)-7/2-2sqrt(3)+2ln(2-sqrt(3))]
=[-1/3-ln(3)/4]-[1/4(-4sqrt(3)+2ln(2-sqrt(3)))]
=[-1/3-ln(3)/4]-[1/4(-4sqrt(3)+ln(4+3-4sqrt(3)))]
=-1/3-ln(3)/4+sqrt(3)-ln(7-4sqrt(3))/4
=-1/3+sqrt(3)-ln(3)/4-ln(7-4sqrt(3))/4
=-1/3+sqrt(3)+1/4ln((7+4sqrt(3))/3)