Question

Evaluate the integral `I = int_(pi/4)^(pi/2) \ 1/sinx \ dx`?

Answer 1

`int_(pi/4)^(pi/2) \ 1/sinx \ dx = ln (sqrt(2) +1) = 0.88137` (5dp)

Explanation:

We want to evaluate:

`I = int_(pi/4)^(pi/2) \ 1/sinx \ dx`
`\ \ = int_(pi/4)^(pi/2) \ cscx \ dx`

The integral is a standard which can be looked up, to get

`I = [-ln|cscx+cotx|]_(pi//4)^(pi//2)`

`\ \ = [ln|cscx+cotx|]_(pi//2)^(pi//4)`

`\ \ = ln abs(csc(pi/4) +cot(pi/4)) - ln abs(csc(pi/2) +cot(pi/2))`

`\ \ = ln abs(sqrt(2) +1) - ln abs(1 +0)`

`\ \ = ln (sqrt(2) +1)`

`\ \ = 0.88137` (5dp)