Question

By using the properties of the definite integral, evaluate the integral:

`int_0^pi log(1+ cos x) dx`

Answer 1

Let `I = int_0^pi log (1 + cos x)  dx`                ....(i)

`I = int_0^pi log [1 + cos (pi - x)] dx`

`[∵ int_0^a f (x) dx = int_0^a f (a - x)  dx]`

`= int_0^pi log (1 - cos x) dx`                              .....(ii)

Adding (i) and (ii), we get

`2 I = int_0^pi [log (1 + cos x) + log (1 - cos x)]  dx`

`= int_0^pi log (1 - cos^2 x)  dx`

`= int_0^pi log sin^2 x  dx`

`= 2 int_0^pilog sin x  dx`

⇒ `I = int_0^pi log sin x  dx`

`= 2 int_0^(pi/2) log sin x  dx = 2I_1`

`[∵ int_0^(2a) f (x) dx = 2 int_0^a f (x) dx, "if" f (2a - x) = f (x)]`

Where `I_1 = int_0^(pi/2) log sin x dx`                ...(iii)

Then, `I_1 = int_0^(pi/2) log sin (pi/2 - x) dx`

⇒ `I_1 = int_0^(pi/2) log cos x dx`                  ....(iv)

Adding (iii) and (iv), we get

`2I_1 = int_0^(pi/2) log sin x dx + int_0^(pi/2) log cos x  dx`

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

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

`= int_0^(pi/2) log ((2sin x cos x)/2)`

`= int_0^(pi/2) log ((sin 2x)/2) dx`

`= int_0^(pi/2) log sin 2 x dx - int_0^(pi/2) log 2 dx`

`= int_0^(pi/2) log sin 2x  dx - (log 2)[x]_0^(pi/2)`

`= int_0^(pi/2) log sin 2 x dx - (log 2) (pi/2 - 0)`

`= int_0^(pi/2) log sin 2x  dx - pi/2 log 2`

`= I_2 - pi/2 log 2`             .....(v)

Where `I_2 = int_0^(pi/2) log sin 2x dx`

Put 2x = t

⇒ 2dx = dt

When x = 0, t = 0

When `x = pi/2, t = pi`

∴ `I_2 = 1/2 int_0^pi log sin t dt`

`= 1/2 *2 int_0^(pi/2) log sin t dt`       `...[∵ log sin (pi -t) = log sint]`

`= int_0^(pi/2) log sin x = I_1`

∴ From (v), we get

`2I_1 = I_2 - pi/2 log 2`

⇒ `2I_1 = I_1 - pi/2 log 2`

⇒ `I_1 = pi/2 log 2`

∴ `I = 2 xx (-pi/2 log 2)`

= - π log 2