If you know the process of finding intsec(theta)d theta∫sec(θ)dθ, this will be very similar.
Very unintuitively, make the modification:
intcsc(theta)d theta=intcsc(theta)((csc(theta)+cot(theta))/(csc(theta)+cot(theta)))d theta=int(csc^2(theta)+csc(theta)cot(theta))/(csc(theta)+cot(theta))d theta∫csc(θ)dθ=∫csc(θ)(csc(θ)+cot(θ)csc(θ)+cot(θ))dθ=∫csc2(θ)+csc(θ)cot(θ)csc(θ)+cot(θ)dθ
Now let u=csc(theta)+cot(theta)u=csc(θ)+cot(θ). Knowing their derivatives, we can say that du=(-csc(theta)cot(theta)-csc^2(theta))d thetadu=(−csc(θ)cot(θ)−csc2(θ))dθ.
Note that the numerator of the integral is just the derivative of its denominator times -1−1.
=-int(-csc(theta)cot(theta)-csc^2(theta))/(csc(theta)+cot(theta))d theta=-int(du)/u=-lnabsu=−∫−csc(θ)cot(θ)−csc2(θ)csc(θ)+cot(θ)dθ=−∫duu=−ln|u|
Reversing the substitution:
intcsc(theta)d theta=-lnabs(csc(theta)+cot(theta))+C∫csc(θ)dθ=−ln|csc(θ)+cot(θ)|+C
