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

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Answer 1 · 2017-06-13T16:36:49

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

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

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

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