Question

Evaluate the following:

`int_0^pi x log sin x "d"x`

Answer 1

Let I = `int_0^pi x log sin x "d"x` ......(i)

= `int_0^pi (pi - x) log sin(pi - x) "d"x`  ....`["Using" int_0^"a" "f"(x)  "d"x = int_0^"a" "f"("a" - x)"d"x]`

I = `int_0^pi (pi - x) log sinx  "d"x`  ......(ii)

Adding (i) and (ii), we get

2I = `int_0^pi [(pi - x) log sin x + x log sinx]"d"x`

2I = `int_0^pi pilog sinx  "d"x`

2I = `2oi int_0^(pi/2) log sinx  "d"x`  ......`[because int_0^"a" "f"(x) "d"x = 2 int_0^("a"/2) "f"(x) "d"x]`

∴ I = `pi int_0^(pi/2) log sinx  "d"x`   .....(iii)

I = `pi int_0^(pi/2) log sin (pi/2 - x) "d"x`

I = `pi int_0^(pi/2) log cos x  "d"x`  ......(iv)

On adding (iii) and (iv), we get

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

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

= `pi int_0^(pi/2)  (log2 sin x cosx)/2  "d"x`

2I = `pi int_0^(pi/2) log sin 2x  "d"x - pi int_0^(pi/2) log 2  "d"x`

Put 2x = t

⇒ 2 dx = dt

⇒ dx = `"dt"/2`

2I = `pi int_0^pi  log sin "t"  "dt" - pi * log 2 int_0^(pi/2)  1 "d"x`  ....[Changing the limit]

2I = `"I" - pi * log 2[x]_0^(pi/2)` ....[From equation (iii)]

2I – I = `- pi^2/2 log 2`

So I = `pi^2/2 log (1/2)`