-0 a ln a -tan x dx- where ae 0- -2 isA- - a ln sin aB- None C- a ln sin aD- - a ln cos a

The correct option is A a ln (sin a)I=∫_0^a ln cos(a x)/sin a. cos xdx = ∫_0^a ln cos x/sin a cos (a x)dx adding 2I=∫_0^aln 1/sin^2 adx = ∫_0^a 2 ln sin a ...

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Byju's Answer

Standard XII

Mathematics

Property 3

∫0 a ln a +ta...

Question

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

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Solution

The correct option is B - a ln (sin a)
$I = \int_{0}^{a} l n \frac{c o s (a - x)}{s i n a . c o s x} d x = \int_{0}^{a} l n \frac{c o s x}{s i n a c o s (a - x)} d x$
adding $2 I = \int_{0}^{a} l n \frac{1}{s i n^{2} a} d x = \int_{0}^{a} - 2 l n s i n a d x = - 2 a l n s i n a$

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∫a0ln(cot a+tan x)dx, where aϵ(0,π2) is

The correct option is B - a ln (sin a)I=∫a0lncos(a−x)sin a.cos xdx=∫a0lncos xsin a cos(a−x)dx adding 2I=∫a0ln1sin2adx=∫a0−2 ln sin a dx=−2a ln sin a

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

The correct option is B - a ln (sin a)
$I = \int_{0}^{a} l n \frac{c o s (a - x)}{s i n a . c o s x} d x = \int_{0}^{a} l n \frac{c o s x}{s i n a c o s (a - x)} d x$
adding $2 I = \int_{0}^{a} l n \frac{1}{s i n^{2} a} d x = \int_{0}^{a} - 2 l n s i n a d x = - 2 a l n s i n a$