Question Number 12926 by kashyappushpendrak1811@gmail.c last updated on 07/May/17
$$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\mathrm{log}\:\mid\mathrm{tan}\:{x}+\mathrm{cot}\:{x}\mid\:{dx}\:= \\ $$
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 .](https://www.tinkutara.com/question/Q12932.png)
$$\mathrm{0}\leqslant{x}\leqslant\pi/\mathrm{2}\:\:\Rightarrow\:\:\:\:\:\:\mathrm{0}\leqslant\mathrm{2}{x}\leqslant\pi \\ $$$$\mathrm{tan}\:{x}+\mathrm{cot}\:{x}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{x}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}{x}} \\ $$$${for}\:\mathrm{0}\leqslant\mathrm{2}{x}\leqslant\pi\:\:\:, \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}{x}}\:\geqslant\mathrm{2}\:. \\ $$$${so},\:\:{I}\:=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\mid\mathrm{tan}\:{x}+\mathrm{cot}\:{x}\mid{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{2}{x}\right){dx} \\ $$$${I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{2}{x}\right){dx}\:\:….\left(\mathrm{1}\right) \\ $$$${I}\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\left[\mathrm{2}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right]{dx}\right. \\ $$$$\:\:{I}\:=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2sec}\:\mathrm{2}{x}\right){dx}\:\:\:…..\left(\mathrm{2}\right) \\ $$$${adding}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right): \\ $$$$\mathrm{2}{I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{4cosec}\:\mathrm{2}{x}.\mathrm{sec}\:\mathrm{2}{x}\:\right){dx} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\frac{\mathrm{4}}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{4}{x}}\right){dx} \\ $$$$\:\:=\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\left[\mathrm{ln}\:\mathrm{4}−\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\:\mathrm{4}{x}\right)\right]{dx} \\ $$$$\:\:=\mathrm{2ln}\:\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}{dx}+\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{4}{x}\right){dx} \\ $$$$\:\:{I}=\left(\mathrm{2ln}\:\mathrm{2}\right)\left(\pi/\mathrm{4}\right)+\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{4}{x}\right){dx} \\ $$$${let}\:\mathrm{2}{x}={z}\:\:\:\Rightarrow\:\:\:{dx}=\frac{{dz}}{\mathrm{2}} \\ $$$${x}=\mathrm{0}\:\:\Rightarrow\:\:{z}=\mathrm{0} \\ $$$${and}\:\:\:{x}=\pi/\mathrm{4}\:\:\:\Rightarrow\:\:{z}=\:\pi/\mathrm{2} \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}\:+\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{2}{z}\right)\frac{{dz}}{\mathrm{2}} \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\left(\mathrm{2cosec}\:\mathrm{2}{x}\right){dx} \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}+\frac{{I}}{\mathrm{2}} \\ $$$${I}\:=\pi\mathrm{ln}\:\mathrm{2}\:. \\ $$