Question

How do you evaluate the integral `int 1/(1-sinx)dx` from 0 to `pi/2`?

Answer 1

The integral is divergent.

Explanation:

This is an inproper integral.

Calculate the indefinite integral first by substitution

Let `u=tan(x/2)`

`=>`, `du=1/2sec^2(x/2)dx=1/2(1+u^2)dx`

And

`sinx=(2u)/(1+u^2)`

Therefore, the integral is

`I=int(dx)/(1-sinx)`

`=int(2/(1+u^2))*1/(1-(2u)/(1+u^2))*du`

`=int(2du)/(1+u^2-2u)`

`=int(2du)/(u-1)^2`

`=-2/(u-1)`

`=-2/(tan(x/2)-1)+C`

The definite integral is

`int_0^(y)(dx)/(1-sinx)=[-2/(tan(x/2)-1)]_0^(y)`

`=(-2/(tan(y/2)-1))-(-2/(0-1))`

`=-2-2/(tan(y/2)-1)`

And the improper integral

`lim_(y->pi/2)(-2-2/(tan(y/2)-1))=-oo`

The integral is divergent.

Answer 2

`int 1/(1-sin x) dx =(tan x + sec x)+ C`
`int _0^(pi/2) 1/(1-sin x) dx =oo`

Explanation:

`int 1/(1-sin x) dx =int (1+sin x)/ (1- sin ^2 x) dx`

`int (1+sin x)/cos ^2 x dx = int (sec^2x +tan x sec x) dx`

`= (tan x + sec x)+ C`

`int _0^(pi/2) 1/(1-sin x) dx = [ tan x + sec x]_0^(pi/2)`

`(oo+oo)-(0+1)= (oo-1) = oo` [Ans]