How do you evaluate the integral #int csctheta#?

Question

Charles Anctil · Accepted Answer

#intcsc(theta)d theta=-lnabs(csc(theta)+cot(theta))+C#

If you know the process of finding #intsec(theta)d theta#, 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#

Now let #u=csc(theta)+cot(theta)#. Knowing their derivatives, we can say that #du=(-csc(theta)cot(theta)-csc^2(theta))d theta#.

Note that the numerator of the integral is just the derivative of its denominator times #-1#.

#=-int(-csc(theta)cot(theta)-csc^2(theta))/(csc(theta)+cot(theta))d theta=-int(du)/u=-lnabsu#

Reversing the substitution:

#intcsc(theta)d theta=-lnabs(csc(theta)+cot(theta))+C#

Adam Adkins · Answer

undefined To evaluate the integral ∫csc(θ) dθ, we can use the substitution method. Let u = csc(θ) and du = -csc(θ) cot(θ) dθ. Thus, the integral becomes ∫-du. Integrating -du yields -u + C, where C is the constant of integration. Finally, substituting back u = csc(θ), the result is -csc(θ) + C. Therefore, the integral of csc(θ) is -csc(θ) + C.