Question

Evaluate:

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

Answer 1

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

Here, `a = π/4` and `b = (3π)/4`, so a + b = π

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

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

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

= `int_(π/4)^((3π)/4) (π - x)/((1 + sin x)) dx`   ....[sin (π − x) = sin x]

Adding, both integrals,

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

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

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

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

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

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

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

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

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

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

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

`2I = π[tanx - secx]_(π/4)^((3π)/4)`

For x = `(3π)/4, tan((3π)/4) = -1, sec((3π)/4) = -sqrt2`

⇒ `tan ((3π)/4) - sec ((3π)/4)`

= `-1 - (-sqrt2)`

= `-1 + sqrt2`

For x = `π/4`

`tan (π/4) = 1, sec (π/4) sqrt2`

⇒ `tan (π/4) - sec (π/4)`

= `1 - sqrt2`

`2I = π[(-1 + sqrt2) - (1 - sqrt2)]`

`2I = π(-1 + sqrt2 - 1 + sqrt2)`

`2I = π (2(sqrt2 - 1))`

`2I = 2π (sqrt2 - 1)`

`I = π(sqrt(2) - 1)`

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