-0a ln a - tan xdx - - where a - 0- -2 is

I = ∫_0^alncos (a x)/sin a cos xdx I = ∫_0^alncos x/sin a cos (a x)dx Adding we get 2I = ∫_0^aln1/(sin a)^2 dx = ∫_0^a 2 ln (sin a) dx = 2a ln (sin a) ⇒ I= ...

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Byju's Answer

Standard XI

Mathematics

Property 3

∫0a ln a + ta...

Question

$\int_{0}^{a}$ $ln (cot a + tan x) d x$ = , where $a ϵ (0, \frac{π}{2})$ is

a ln (sin a)

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- a ln (sin a)

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- a ln (cos a)

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none of these

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Solution

The correct option is B $- a ln (sin a)$
$I = \int_{0}^{a} ln \frac{cos (a - x)}{sin a cos x} d x$

$I = \int_{0}^{a} ln \frac{cos x}{sin a cos (a - x)} d x$
Adding we get $2 I = \int_{0}^{a} ln \frac{1}{(sin a)^{2}} d x$
$= \int_{0}^{a} - 2 ln (sin a) d x = - 2 a ln (sin a) \Rightarrow I = - a ln (sin a)$

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∫a0 ln(cota+tanx)dx = , where aϵ(0,π2) is

The correct option is B –aln(sina)I=∫a0lncos(a−x)sinacosxdx I=∫a0lncosxsinacos(a−x)dx Adding we get 2I=∫a0ln1(sina)2dx =∫a0−2ln(sina)dx=−2aln(sina)⇒I=−aln(sina)

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