-0aln a-tan x d x- where a -0- -2 is

The correct option is B a In (sin a) I=∫_0^aIn cos (a x)/sin a.cos xdx =∫_0^aIn cos x/sin a cos (a x)dx Adding 2I =∫_0^aIn 1/sin^2 adx =∫_0^a 2 Insin a dx = 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Property 3

∫0aln a+tan x...

Question

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

a in (sin a)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-a In (sin a)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

– a In (cos a)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
-a In (sin a)

$I = \int_{0}^{a} I n \frac{c o s (a - x)}{s i n a . c o s x} d x = \int_{0}^{a} I n \frac{c o s x}{s i n a c o s (a - x)} d x$
Adding $2 I = \int_{0}^{a} I n \frac{1}{s i n^{2} a} d x = \int_{0}^{a} - 2 I n s i n a d x = - 2$ a In sin a

Suggest Corrections

Similar questions

\int_{0}^{a}

ln (cot a + tan x) d x

= , where

a ϵ (0, \frac{π}{2})

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

, where

a ϵ (0, \frac{π}{2})

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

Q. Evaluate:

\int {tan}^{- 1} (sec x + tan x) d x

where

- \frac{π}{2} < x < \frac{π}{2}

\int_{0}^{a}

ln (cot a + tan x) d x

= , where

a ϵ (0, \frac{π}{2})

Join BYJU'S Learning Program

∫a0 ln(cot a +tan x)dx, where a ϵ(0,π2) is

The correct option is B -a In (sin a) I=∫a0Incos(a−x)sina.cosxdx=∫a0Incosxsinacos(a−x)dx Adding 2I=∫a0In1sin2adx=∫a0−2Insinadx=−2 a In sin a

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

The correct option is B
-a In (sin a)

$I = \int_{0}^{a} I n \frac{c o s (a - x)}{s i n a . c o s x} d x = \int_{0}^{a} I n \frac{c o s x}{s i n a c o s (a - x)} d x$
Adding $2 I = \int_{0}^{a} I n \frac{1}{s i n^{2} a} d x = \int_{0}^{a} - 2 I n s i n a d x = - 2$ a In sin a