Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 53

int1/(2-3sin(theta))d theta
Apply integral substitution:u=tan(theta/2)
=>du=1/2sec^2(theta/2)d theta
Use the trigonometric identity:sec^2(x)=1+tan^2(x)
sec^2(theta/2)=1+tan^2(theta/2)
sec^2(theta/2)=1+u^2
du=1/2(1+u^2)d theta
d theta=2/(1+u^2)du
From integral substitution:u=tan(theta/2)
=>sin(theta/2)=u/sqrt(u^2+1)
cos(theta/2)=1/sqrt(u^2+1)
sin(theta)=2sin(theta/2)cos(theta/2)
sin(theta)=2(u/sqrt(u^2+1))(1/sqrt(u^2+1))
sin(theta)=(2u)/(u^2+1)
Now the integrand can be written as :
int1/(2-3sin(theta))d theta=int1/(2-3((2u)/(u^2+1)))(2/(1+u^2))du
=int1/((2(u^2+1)-3(2u))/(u^2+1))(2/(1+u^2))du
=int2/(2u^2+2-6u)du
=int2/(2(u^2-3u+1))du
=int1/(u^2-3u+1)du
Complete the square of the denominator,
=int1/((u-3/2)^2-5/4)du
Again apply integral substitution:v=u-3/2
=>dv=1du
=int1/(v^2-(sqrt(5)/2)^2)dv
=int1/(-1((sqrt(5)/2)^2-v^2))dv
Take the constant out,
=-1int1/((sqrt(5)/2)^2-v^2)dv
Now use the standard table integral:int1/(a^2-x^2)dx=1/(2a)ln|(a+x)/(a-x)|+C
=-1(1/(2(sqrt(5)/2))ln|(sqrt(5)/2+v)/(sqrt(5)/2-v)|)+C
=(-1/sqrt(5))ln|(sqrt(5)+2v)/(sqrt(5)-2v)|+C
Substitute back v=u-3/2
=(-1/sqrt(5))ln|(sqrt(5)+2(u-3/2))/(sqrt(5)-2(u-3/2))|+C
=(-1/sqrt(5))ln|(sqrt(5)+2u-6)/(sqrt(5)-2u+6)|+C
Substitute back u=tan(theta/2)
=(-1/sqrt(5))ln|(sqrt(5)+2tan(theta/2)-6)/(sqrt(5)-2tan(theta/2)+6)|+C