Question

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

Options

• `pi/4 + 2/3 tan^-1 3`

• `- pi/3 - 2/3 tan^-1 3`

• `- pi/12 + 2/3 tan^-1 2`

• `pi/6 - 2/3 tan^-1 4`

Answer 1

`bb(- pi/12 + 2/3 tan^-1 2)`

Explanation:

We have, `int_0^(pi/4) (cos^2 x)/(cos^2 x + 4 sin^2 x) dx`

= `int_0^(pi/4) (dx)/(1 + 4 tan^2 x)`

Let tan x = t

= sec² x dx = dt

= `dx = (dx)/(sec^2 x) = (dt)/(1 + t^2)`   ...(i)

= when x = 0, t = 0, and when x = `pi/4`, t = 1

From Eq. (i), `int_0^1 (dt)/((1 + t^2)(1 + 4t^2))`

= `1/3 int_0^1 (4(1 + t^2) - (1 + 4t^2))/((1 + 4t^2)(1 + t^2))dx`

= `1/3 int_0^1 (4/(1 + 4t^2) - 1/(1 + t^2)dt)`

= `1/3 [4/4 int_0^1 (dt)/((1/2)^2 + t^2) - int_0^1 1/(1 + t^2)dt]`

= `1/3 [1/(1"/"2) (tan^-1  t/(1"/"2))_0^1 - (tan^-1 t)_0^1]`

= `1/3 [2 (tan^-1 2t)_0^1 - pi/4]`

= `1/3 [2 tan^-1 (2) - pi/4]`

= `-pi/12 + 2/3 tan^-1 2`