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




Question Number 12926 by kashyappushpendrak1811@gmail.c last updated on 07/May/17
∫_( 0) ^(π/2)  log ∣tan x+cot x∣ dx =
$$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\mathrm{log}\:\mid\mathrm{tan}\:{x}+\mathrm{cot}\:{x}\mid\:{dx}\:= \\ $$
Answered by ajfour last updated on 07/May/17
0≤x≤π/2  ⇒      0≤2x≤π  tan x+cot x=((sin x)/(cos x))+((cos x)/(sin x))  =(1/((sin x)(cos x))) = (2/(sin 2x))  for 0≤2x≤π   ,  (2/(sin 2x)) ≥2 .  so,  I =∫_0 ^(π/2) ln ∣tan x+cot x∣dx             = ∫_0 ^(π/2) ln (2cosec 2x)dx  I=2∫_0 ^(π/4) ln (2cosec 2x)dx  ....(1)  I = 2∫_0 ^(π/4) ln (2cosec [2((π/4)−x)]dx    I =2∫_0 ^(π/4) ln (2sec 2x)dx   .....(2)  adding (1) & (2):  2I=2∫_0 ^(π/4) ln (4cosec 2x.sec 2x )dx  I=∫_0 ^(π/4) ln ((4/((1/2)sin 4x)))dx    =∫_0 ^(π/4) [ln 4−ln ((1/2)sin  4x)]dx    =2ln 2∫_0 ^(π/4) dx+∫_0 ^(π/4) ln (2cosec 4x)dx    I=(2ln 2)(π/4)+∫_0 ^(π/4) ln (2cosec 4x)dx  let 2x=z   ⇒   dx=(dz/2)  x=0  ⇒  z=0  and   x=π/4   ⇒  z= π/2  I =(π/2)ln 2 +∫_0 ^(π/2) ln (2cosec 2z)(dz/2)  I =(π/2)ln 2+(1/2)∫_0 ^(π/2) ln (2cosec 2x)dx  I =(π/2)ln 2+(I/2)  I =πln 2 .
$$\mathrm{0}\leqslant{x}\leqslant\pi/\mathrm{2}\:\:\Rightarrow\:\:\:\:\:\:\mathrm{0}\leqslant\mathrm{2}{x}\leqslant\pi \\ $$$$\mathrm{tan}\:{x}+\mathrm{cot}\:{x}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{x}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}{x}} \\ $$$${for}\:\mathrm{0}\leqslant\mathrm{2}{x}\leqslant\pi\:\:\:, \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}{x}}\:\geqslant\mathrm{2}\:. \\ $$$${so},\:\:{I}\:=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\mid\mathrm{tan}\:{x}+\mathrm{cot}\:{x}\mid{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{2}{x}\right){dx} \\ $$$${I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{2}{x}\right){dx}\:\:….\left(\mathrm{1}\right) \\ $$$${I}\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\left[\mathrm{2}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right]{dx}\right. \\ $$$$\:\:{I}\:=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2sec}\:\mathrm{2}{x}\right){dx}\:\:\:…..\left(\mathrm{2}\right) \\ $$$${adding}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right): \\ $$$$\mathrm{2}{I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{4cosec}\:\mathrm{2}{x}.\mathrm{sec}\:\mathrm{2}{x}\:\right){dx} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\frac{\mathrm{4}}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{4}{x}}\right){dx} \\ $$$$\:\:=\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\left[\mathrm{ln}\:\mathrm{4}−\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\:\mathrm{4}{x}\right)\right]{dx} \\ $$$$\:\:=\mathrm{2ln}\:\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}{dx}+\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{4}{x}\right){dx} \\ $$$$\:\:{I}=\left(\mathrm{2ln}\:\mathrm{2}\right)\left(\pi/\mathrm{4}\right)+\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{4}{x}\right){dx} \\ $$$${let}\:\mathrm{2}{x}={z}\:\:\:\Rightarrow\:\:\:{dx}=\frac{{dz}}{\mathrm{2}} \\ $$$${x}=\mathrm{0}\:\:\Rightarrow\:\:{z}=\mathrm{0} \\ $$$${and}\:\:\:{x}=\pi/\mathrm{4}\:\:\:\Rightarrow\:\:{z}=\:\pi/\mathrm{2} \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}\:+\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{2}{z}\right)\frac{{dz}}{\mathrm{2}} \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{2}{x}\right){dx} \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}+\frac{{I}}{\mathrm{2}} \\ $$$${I}\:=\pi\mathrm{ln}\:\mathrm{2}\:. \\ $$

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