Question

Evaluate: 

`int_0^(pi/4) (x dx)/(1+cos2x+sin2x)`

Answer 1

`int_0^(pi/4) (x dx)/(1+cos2x+sin2x)`

by using property '4' we get

`I = int_0^(pi/4) ((pi/4 - x) dx)/(1 + cos (pi/2 - 2x) + sin (pi/2 - 2x))`

`I = int_0^(pi/4) ((pi/4 - x) dx)/(1+sin 2x + cos2x)`

Now add eq (i) and (ii)

`2I = int_0^(pi/4) (pi/4)/(1+sin 2x + cos2x)  dx`

`2I = pi/4 int_0^(pi/4) (dx)/(2 cos^2x + 2sinx cos2x)`     ...[∴ 1 + cos 2 θ = 2 cos² θ, sin 2 θ = 2 sin θ cos θ]

⇒ `2I = pi/8 int_0^(pi/4) (1/cos^2x)/(1+tanx)`

⇒ `I = pi/16 int_0^(pi/4) (sec^2x)/(1+tanx) dx`

Let 1 + tan x = t

sec²x dx = dt

New limit

1 + tan 0 = t ⇒ t = 1

`1 + tan  pi/4t => t = 2`

⇒ `I = pi/16 int_1^2 dt/t`

⇒ `I = pi/16 |log t|_1^2`

`I = pi/16 log(2) - log(1)`

`I = pi/16 log2`