I -0 a ln a -tan x dx- where a -0- -2- then I is equal toA- - a ln sin a B- 2 a ln cos aC- -2 a ln sinaD- a ln sina

The correct option is A –a ln(sin a) I = ∫^a_0 ln (cot a + tan x)dx = ∫^a_0 ln (cos(a x)/sin a cos x) dx ...(1) ∴ I = ∫^a_0 ln (cos x/sin a cos (a ...

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Byju's Answer

Standard XII

Mathematics

Property 3

I =∫0 a ln a ...

Question

$I = \int_{0}^{a} l n (c o t a + t a n x) d x$ , where $a ϵ (0, \frac{π}{2})$ , then I is equal to

$a ln (s i n a)$

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$- a ln (s i n a)$

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$2 a ln (c o s a)$

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$- 2 a ln (s i n a)$

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Solution

The correct option is B
$- a ln (s i n a)$

$I = \int_{0}^{a} l n (c o t a + t a n x) d x$

$= \int_{0}^{a} l n (\frac{c o s (a - x)}{s i n a c o s x}) d x$ ...(1)

$∴ I = \int_{0}^{a} l n (\frac{c o s x}{s i n a c o s (a - x)}) d x$ ...(2)

Adding (1) and (2) we get $2 I = \int_{0}^{a} l n (\frac{1}{s i n^{2} a}) d x$

$2 I = - 2 \int_{0}^{a} l n (s i n a) d x$

$I = - a l n (s i n a)$

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I=∫a0ln(cot a+tan x)dx, where aϵ(0,π2), then I is equal to

The correct option is B –aln(sina) I=∫a0ln(cot a+tan x)dx =∫a0ln(cos(a−x)sin a cos x)dx ...(1) ∴I=∫a0ln(cos xsin a cos (a−x))dx ...(2) Adding (1) and (2) we get 2I=∫a0ln(1sin2a)dx 2I=−2∫a0ln(sin a)dx I=−a ln (sin a)

$I = \int_{0}^{a} l n (c o t a + t a n x) d x$ , where $a ϵ (0, \frac{π}{2})$ , then I is equal to