Question

Evaluate: `int_0^pi 1/(3 + 2sinx + cosx)  "d"x`

Answer 1

Let I = `int_0^pi 1/(3 + 2sinx + cosx)  "d"x`

Put `tan (x/2)` = t

∴ x = 2tan⁻¹t

∴ dx = `(2"dt")/(1 + "t"^2)`, sin x = `(2"t")/(1 + "t"^2)` and cos x = `(1 - "t"^2)/(1 + "t"^2)`

When x = 0, t = 0 and when x = π, t = ∞

∴ I = `int_0^∞ 1/(3 + 2((2"t")/(1 + "t"^2)) + (1 - "t"^2)/(1 + "t"^2)) xx (2  "dt")/(1 + "t"^2)`

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

= `int_0^∞ (2  "dt")/(2"t"^2 + 4"t" + 4)`

= `int_0^∞  "dt"/("t"^2 + 2"t" + 2)`

= `int_0^∞  "dt"/("t"^2 + 2"t" + 1 + 1)`

= `int_0^∞  "dt"/(("t" + 1)^2 + 1^2)`

= `[tan^-1 ("t" + 1)]_0^∞`

= `tan^-1(1 + ∞) - tan^-1(1 + 0)`

= `tan^-1(∞) - tan^-1 (1)`

= `pi/2 - pi/4`

∴ I = = `pi/4`