Question

Evaluate : `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x +1)*dx`

Answer 1

Let I = `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x +1)*dx`

Put tan x = t
∴ sec²x·dx = dt
When x = 0, t = tan 0 = 0

When x = `pi/(4), t = tan  pi/(4)` = 1

∴ I = `int_0^1 dt/(3t^2 + 4t + 1)`

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

= `(1)/(3) int_0^1 dt/(t^2 + (4t)/(3) + (4)/(9) - (4)/(9) + (1)/(3)`

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

= `(1)/(3)(1)/(2(1/3))[log |(t + 2/3 - 1/3)/(t + 2/3 + 1/3)|]_0^1`

= `(1)/(2)[log ((1 + 1/3)/(1 + 1)) - log((0 + 1/3)/(0 + 1))]`

= `(1)/(2)[log (2/3) - log(1/3)`

= `(1)/(2)log2`.