Question

How do you find the integral of `int cscx dx` from pi/2 to pi?

Answer 1

Integral is divergent

Explanation:

This is an interesting intergal, we can find its antiderivative via considering a valid substitution.

`intcscx dx`

let `u` = `cotx`
then `du = -csc^2 x dx`

Via quotient rule;
`cotx` = `cosx/sinx`
hence `d/dx(cotx)` = `((sinx)(-sinx)-(cosx)(cosx))/sin^2 x`
Hence = `-csc^2 x`

Hence `(-du)/cscx = cscx dx`

Hence `intcscx dx` becomes `-int (du)/cscx`

Considering `1 + cot^2 x = csc^2 x`

Hence if `u` = `cotx` then `cscx` = `(1+u^2)^(1/2)`

Hence `-int (du)/cscx` becomes `-int (du)/(1+u^2)^(1/2)`

Then make a new substitution of `u` = `sinhtheta`
Hence `du` = `coshtheta d theta`

Now by cosindering `cosh^2 theta - sinh^2 theta = 1`

`-int (du)/(1+u^2)^(1/2)` becomes `-int d theta`

= `-theta + c`

As `u` = `sinh theta` then `theta` = `arcsinh(u)`

Hence `-arcsinh (u) + c`

Hence as `u` = `cottheta`

Hence `int csc xdx` = `c - sinh^-1(cotx)`

So hence we evaluate the antiderivative from `pi/2` to `pi`, but there is issue in this as `sinh^-1( cot(pi) )` is undifined