Question

How do you evaluate the integral `int csctheta`?

Answer 1

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

Explanation:

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`