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By using the properties of the definite integral, evaluate the integral: ∫0π2(2logsinx-logsin2x)dx

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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`

योग
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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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अध्याय 7: Integrals - Exercise 7.11 [पृष्ठ ३४७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.11 | Q 10 | पृष्ठ ३४७

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