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प्रश्न
Evaluate the following:
`int_0^pi x log sin x "d"x`
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उत्तर
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)`
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