-0-2log tan x- x d x-

The correct option is B π log 2∫_0^π/2 log (tan x+cot x)dx=∫_0^π/2 log [ 2/sin 2x ]dx = ∫_0^π/2(log 2 log sin 2x)dx=log 2 ( π/2 )+( π/2 )log 2=π log 2 ...

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

Byju's Answer

Standard XII

Mathematics

Geometric Interpretation of Def.Int as Limit of Sum

∫0π/2log tan ...

Question

$\int_{0}^{\frac{π}{2}} l o g (t a n x + c o t x) d x =$

π l o g 2

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

- π l o g 2

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

- \frac{π}{2} l o g 2

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

\frac{π}{2} l o g 2

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

Open in App

Solution

The correct option is A $π l o g 2$
$\int_{0}^{\frac{π}{2}} l o g (t a n x + c o t x) d x = \int_{0}^{\frac{π}{2}} l o g [\frac{2}{s i n 2 x}] d x$
$= \int_{0}^{\frac{π}{2}} (l o g 2 - l o g s i n 2 x) d x = l o g 2 (\frac{π}{2}) + (\frac{π}{2}) l o g 2 = π l o g 2$

Suggest Corrections

Similar questions

\int_{0}^{π / 2} log (tan x + cot x) d x =

\int_{0}^{π / 2} log (tan x + cot x) d x = π (log 2)

Q. Solve:

\int_{0}^{π / 4} {tan}^{100} x d x + \int_{0}^{π / 4} {tan}^{102} x d x = . . . .

Q. If

I_{1} = \int_{0}^{π / 4} {sin}^{2} x d x

and

I_{2} = \int_{0}^{π / 4} {cos}^{2} x d x

, then

\int_{0}^{π / 2} \frac{1}{1 + \cot x} d x

Join BYJU'S Learning Program

∫π20 log(tan x+cot x)dx=

The correct option is A π log 2∫π20 log(tan x+cot x)dx=∫π20 log[2sin 2x]dx =∫π20(log2−log sin 2x)dx=log 2(π2)+(π2)log 2=π log 2

$\int_{0}^{\frac{π}{2}} l o g (t a n x + c o t x) d x =$

The correct option is A $π l o g 2$
$\int_{0}^{\frac{π}{2}} l o g (t a n x + c o t x) d x = \int_{0}^{\frac{π}{2}} l o g [\frac{2}{s i n 2 x}] d x$
$= \int_{0}^{\frac{π}{2}} (l o g 2 - l o g s i n 2 x) d x = l o g 2 (\frac{π}{2}) + (\frac{π}{2}) l o g 2 = π l o g 2$