結果 : integration of log cosx limit 0 to pi 2
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integration 0 to 'pi/2' ln(cosx)dx || integration 0 to 'pi/2' log(sinx)dx || 0 to pi/2 ln(cosx)dx

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lntegration of log cosx.... solution using properties of definite integral

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integrate [log(sin x) - log(2cos x)] dx from 0 to pi/2

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`int_0 ^ (pi/2) log(cosx) dx=`

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Prove that: `int_(0)^(pi//2) log (sin x) dx =int_(0)^(pi//2) log (cos x) dx =(-pi)/(2) log 2`

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Show that 0 to pi/2 integral x/sinx+cosx dx = π/2√2 log (√2+1) || Vvvip || Definite Integration

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`int_0 ^(pi/2) log sinx dx = int_0 ^(pi/2) log cosx dx = 1/2 (pi) log(1/2)`

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Prove that integral of cos square x upon sinx plus cosx dx from 0 to pi by 2 equal 1/√2 log(√2 + 1)

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