0-pi-2-log-tan-x-cot-x-dx-

Question Number 12926 by kashyappushpendrak1811@gmail.c last updated on 07/May/17

\underset{0}{\int^{π / 2}} \log ∣ \tan x + \cot x ∣ d x =

Answered by ajfour last updated on 07/May/17

0 ⩽ x ⩽ π / 2 \Rightarrow 0 ⩽ 2 x ⩽ π

\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}

= \frac{1}{(\sin x) (\cos x)} = \frac{2}{\sin 2 x}

f o r 0 ⩽ 2 x ⩽ π,

\frac{2}{\sin 2 x} ⩾ 2 .

s o, I = \underset{0}{\int^{π / 2}} \ln ∣ \tan x + \cot x ∣ d x

= \underset{0}{\int^{π / 2}} \ln (2 cosec 2 x) d x

I = 2 \underset{0}{\int^{π / 4}} \ln (2 cosec 2 x) d x \dots . (1)

I = 2 \underset{0}{\int^{π / 4}} \ln (2 cosec [2 (\frac{π}{4} - x)] d x

I = 2 \underset{0}{\int^{π / 4}} \ln (2 \sec 2 x) d x \dots . . (2)

a d d i n g (1) & (2) :

2 I = 2 \underset{0}{\int^{π / 4}} \ln (4 cosec 2 x . \sec 2 x) d x

I = \underset{0}{\int^{π / 4}} \ln (\frac{4}{\frac{1}{2} \sin 4 x}) d x

= \underset{0}{\int^{π / 4}} [\ln 4 - \ln (\frac{1}{2} \sin 4 x)] d x

= 2 \ln 2 \underset{0}{\int^{π / 4}} d x + \underset{0}{\int^{π / 4}} \ln (2 cosec 4 x) d x

I = (2 \ln 2) (π / 4) + \underset{0}{\int^{π / 4}} \ln (2 cosec 4 x) d x

l e t 2 x = z \Rightarrow d x = \frac{d z}{2}

x = 0 \Rightarrow z = 0

a n d x = π / 4 \Rightarrow z = π / 2

I = \frac{π}{2} \ln 2 + \underset{0}{\int^{π / 2}} \ln (2 cosec 2 z) \frac{d z}{2}

I = \frac{π}{2} \ln 2 + \frac{1}{2} \underset{0}{\int^{π / 2}} \ln (2 cosec 2 x) d x

I = \frac{π}{2} \ln 2 + \frac{I}{2}

I = π \ln 2 .

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