`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`