1/2
s log
a + b sin x
a + bcos x
dx =
0
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Answer:I = (π/2)∫0 log[(a + bsinx)/(a + b cos x)]dx .....(1) = (π/2)∫0 log[{a + b sin[(π/2) – x]}/{a + b cos[(π/2) – x]}]dx ∴ I = (π/2)∫0 log[(a + bcosx)/(a + b sin x)]dx ....(2) (1) + (2) given, 2I = (π/2)∫0 log[(a + bsinx)/(a + bcosx)] + log[(a + bcosx)/(a + bsinx)] ∙dx 2I = (π/2)∫0 log(1)dx ∴ 2I = 0 I = 0.
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