Question

Prove:

`int_(π/4)^((3π)/4) (xdx)/(1 + sinx) = (sqrt(2) - 1)π`

Answer 1

Given: `int_(π/4)^((3π)/4) (xdx)/(1 + sinx)`

To Prove: `(sqrt(2) - 1)π`

Proof:

`I = int_(π/4)^((3π)/4) (xdx)/(1 + sinx)`   ...(i)

By applying property of definite integral

`int_a^b f(x) dx = int_a^b f(a + b - x) dx`

`I = int_(π/4)^((3π)/4) (((3π)/4 + π/4 - x)dx)/(1 + sin((3π)/4 + π/4 - x)`

⇒ `I = int_(π/4)^((3π)/4) ((π - x)dx)/(1 + sin x)`   ...(ii)

On adding (i) and (ii)

`2I = int_(π/4)^((3π)/4) (x + (π - x))/(1 + sin x) dx`

⇒ `2I = π int_(π/4)^((3π)/4) dx/(1 + sin x)`

⇒ `I = π/2 int_(π/4)^((3π)/4) 1/(1 + sin x) xx (1 - sin x)/(1 - sin x) dx`

⇒ `I = π/2 int_(π/4)^((3π)/4) (1 - sin x)/(1 - sin^2x) dx`

⇒ `I = π/2 int_(π/4)^((3π)/4) (1 - sin x)/(cos^2 x) dx`

⇒ `I = π/2 int_(π/4)^((3π)/4)(sec^2 x - tan x sec x)dx`

⇒ `I = π/2 [tan x - sec x]_(π/4)^((3π)/4)`

⇒ `I = π/2 [(tan  (3π)/4 - sec  (3π)/4) - (tan  π/4 - sec  π/4)]`

⇒ `I = π/2 [(-1 + sqrt(2)) - (1 - sqrt(2))]`

= `π/2 [2sqrt(2) - 2]`

= `π(sqrt(2) - 1)`

Hence Proved.