Question

How do you evaluate `int (1/(sinx)^2)` from -pi/4 to pi/4?

Answer 1

Because the integrand is not defined at `x=0`, this is an improper integral.

Explanation:

`1/(sinx)^2sinx = (1/sinx)^2 = csc^2x`

`int_(-pi/4)^(pi/4) csc^2x dx = lim_(brarr0^-)int_(-pi/4)^b csc^2x dx + lim_(ararr0^+)int_a^(pi/4) csc^2x dx`,
provided that

Thinking about the derivtaives of trigonometric functions, we should recall that `d/dx(cotx) = -csc^2x`.

So `int 1/(sinx)^2 dx = -cotx`

As `xrarr0`, `cotx rarr+- oo` so the integrals on the right do not converge, so the main integral does not converge.