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Question
Evaluate the following : `int_0^(pi/2) [2 log (sinx) - log (sin 2x)]*dx`
Solution
Let I = `int_0^(pi/2) (2 log sinx - log sin 2x)*dx`
= `int_0^(pi/2) [2log sinx - log (2sinx cosx)]*dx`
= `int_0^(pi/2) [2log sinx - (log 2 + log sinx + log cosx)]*dx`
= `int_0^(pi/2) (2 log sinx - log 2 - log sinx - log cos x)*dx`
= `int_0^(pi/2) (log sinx - log cosx - log 2)*dx`
= `int_0^(pi/2) log sinx*dx - int_0^(pi/2) log cosx*dx - log2 int_0^(pi/2) 1*dx`
= `int_0^(pi/2) log [sin(pi/2 - x)]*dx - int_0^(pi/2) logcosx*dx - log2[x]_0^(pi/2) ...[because int_0^a f(x)*dx = iint_0^a f(a - x)*dx]`
= `int_0^(pi/2) logcosx*dx - int_0^(pi/2) logcosx*dx - log2[pi/2 - 0]`
= `- pi/(2) log 2`.
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