Question

By using the properties of the definite integral, evaluate the integral:

`int_0^pi (x  dx)/(1+ sin x)`

Answer 1

Let `I = int_0^pi (x  dx)/ (1 + sin x)`

`I = int_0^pi (pi - x)/ (1 + sin (pi-x)) dx`           `...[∵ int_0^a f (x) dx = int_0^a f (a - x) dx]`

`= int_0^pi (pi - x)/ (1 + sin x)  dx`

Adding (i) and (ii), we get

`2 I = int_0^pi (x + pi - x)/ (1 + sin x)  dx`

`= pi int_0^pi 1/ (1 +  sin x)  dx`

`= pi int_0^pi (1 - sin x)/ (1 - sin^2 x)  dx`

`= pi int_0^pi (1 - sin x)/(cos^2 x) dx`

`= pi int_0^pi (sec^2 x - tan x sec x) dx`

`= pi [tan x - sec x]_0^pi`

= π [(tan π - sec π) - (tan0 - sec0)]

= π [(0 - (-1)) - (0 - 1)]

= 2π

Hence, I = π