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प्रश्न
By using the properties of the definite integral, evaluate the integral:
`int_0^(pi/2) (2log sin x - log sin 2x)dx`
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उत्तर
Let `I = int_0^(pi/2) (2 log sin x - log sin 2x) dx`
`= int_0^(pi/2) [2 log sin x - log (2 sin x cos x)] dx`
`= int_0^(pi/2) [2 log sinx - log 2 - log sin x - log cos x] dx`
`= int_0^(pi/2) [log sin x - log 2 - log cos x] dx`
`= int_0^(pi/2) log sin x dx - int_0^(pi/2) log 2 dx - int_0^(pi/2) log cos x dx`
`= int_0^(pi/2) log sin x dx - int_0^(pi/2) log 2 dx - int_0^(pi/2) log cos (pi/2 - x) dx` `....[∵ int_0^a f (x) dx = int_0^a f (a - x) dx]`
`= int_0^(pi/2) log sinx dx - (log 2) [x]_0^(pi/2) - int_0^(pi/2) log sin x dx`
`= - (log 2) (pi/2 - 0)`
`= pi/2 log2`
`= pi/2 log (2)^-1`
`= pi/2 log (1/2)`
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