Question

Evaluate: `int_0^π 1/(5 + 4 cos x)dx`

Options

• `π/3`

• `π/2`

• `π/4`

• `π/6`

Answer 1

`bb(π/3)`

Explanation:

We have, I = `int_0^π 1/(5 + 4cosx)dx`

= `int_0^π 1/(5 + 4((1 - tan^2  x/2)/(1 + tan^2  x/2))) dx`

= `int_0^π (1 + tan^2  x/2)/(5(1 + tan^2  x/2) + 4(1 - tan^2  x/2))dx`

= `int_0^π (1 + tan^2  x/2)/(9 + tan^2  x/2)dx`

= `int_0^π (sec^2  x/2)/(9 + tan^2  x/2)dx`

Let `tan  x/2` = t

`\implies 1/2 sec^2  x/2dx` = dt

Also, x = 0

`\implies` t = 0 and x = π

`\implies` t = ∞

∴ I = `int_0^∞ (dt)/(9 + t^2)`

∴ I = `2int_0^∞ (dt)/(3^2 + t^2)`

∴ I = `2/3[tan^-1  t/3]_0^∞`

= `2/3[tan^-1∞ - tan^-1 0]`

= `2/3(π/2 - 0)`

= `π/3`