Question Number 182045 by Emrice last updated on 03/Dec/22
$$ \\ $$$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left({tanx}\right)^{\mathrm{2013}} }\:{dx}\:=\:? \\ $$$$ \\ $$
Commented by Frix last updated on 03/Dec/22
$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{{n}} \:{x}}\:\mathrm{is}\:\mathrm{symmetric}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point}\:\left(\frac{\pi}{\mathrm{4}};\:\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\underset{\frac{\pi}{\mathrm{4}}−{a}} {\overset{\frac{\pi}{\mathrm{4}}+{a}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{tan}^{{n}} \:{x}}={a}\forall{n}\in\mathbb{N}\:\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\frac{\pi}{\mathrm{12}} \\ $$
Answered by Ar Brandon last updated on 03/Dec/22
![I=∫_(π/6) ^(π/3) (1/(1+(tanx)^(2013) ))dx I=∫_(π/6) ^(π/3) (((cosx)^(2013) )/((cosx)^(2013) +(sinx)^(2013) ))dx ...eqn(i) I=∫_(π/6) ^(π/3) (((sinx)^(2013) )/((cosx)^(2013) +(sinx)^(2013) ))dx ...eqn(ii) eqn(i)+eqn(ii) 2I=∫_(π/6) ^(π/3) (((cosx)^(2013) +(sinx)^(2013) )/((cosx)^(2013) +(sinx)^(2013) ))dx=∫_(π/6) ^(π/3) dx ⇒I=(1/2)[x]_(π/6) ^(π/3) =(1/2)((π/3)−(π/6))=(1/2)((π/6))= determinant (((π/(12))))](https://www.tinkutara.com/question/Q182061.png)
$${I}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{tan}{x}\right)^{\mathrm{2013}} }{dx} \\ $$$${I}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\left(\mathrm{cos}{x}\right)^{\mathrm{2013}} }{\left(\mathrm{cos}{x}\right)^{\mathrm{2013}} +\left(\mathrm{sin}{x}\right)^{\mathrm{2013}} }{dx}\:\:\:…\mathrm{eqn}\left({i}\right) \\ $$$${I}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\left(\mathrm{sin}{x}\right)^{\mathrm{2013}} }{\left(\mathrm{cos}{x}\right)^{\mathrm{2013}} +\left(\mathrm{sin}{x}\right)^{\mathrm{2013}} }{dx}\:\:\:…\mathrm{eqn}\left({ii}\right) \\ $$$$\mathrm{eqn}\left({i}\right)+\mathrm{eqn}\left({ii}\right) \\ $$$$\mathrm{2}{I}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\left(\mathrm{cos}{x}\right)^{\mathrm{2013}} +\left(\mathrm{sin}{x}\right)^{\mathrm{2013}} }{\left(\mathrm{cos}{x}\right)^{\mathrm{2013}} +\left(\mathrm{sin}{x}\right)^{\mathrm{2013}} }{dx}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} {dx} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\left[{x}\right]_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}\right)=\begin{array}{|c|}{\frac{\pi}{\mathrm{12}}}\\\hline\end{array} \\ $$
Answered by SEKRET last updated on 04/Dec/22
$$\:\:\int_{\boldsymbol{\mathrm{a}}} ^{\boldsymbol{\mathrm{b}}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}}=\int_{\boldsymbol{\mathrm{a}}} ^{\boldsymbol{\mathrm{b}}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\:\:\int_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{3}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{\mathrm{tanx}}\right)^{\mathrm{2013}} }\boldsymbol{\mathrm{dx}}=\:\:\boldsymbol{\mathrm{I}} \\ $$$$\:\:\:\int_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{3}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{\mathrm{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{3}}+\frac{\boldsymbol{\pi}}{\mathrm{6}}−{x}\right)\right)^{\mathrm{2013}} }\boldsymbol{\mathrm{dx}}= \\ $$$$=\int_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\frac{\boldsymbol{\pi}}{\mathrm{3}}} \frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{tanx}}\right)^{\mathrm{2013}} }}\:\boldsymbol{\mathrm{dx}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\:\frac{\pi}{\mathrm{3}}} \frac{\:\left(\boldsymbol{\mathrm{tanx}}\right)^{\mathrm{2013}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{\mathrm{tanx}}\right)^{\mathrm{2013}} }\boldsymbol{\mathrm{dx}}= \\ $$$$\:\:\int_{\frac{\pi}{\mathrm{6}}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{3}}} \mathrm{1}\:\boldsymbol{\mathrm{dx}}\:−\int_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{3}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{\mathrm{tanx}}\right)^{\mathrm{2013}} }\boldsymbol{\mathrm{dx}}=\:\boldsymbol{\mathrm{I}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{x}}\:\:/_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\:\:\frac{\boldsymbol{\pi}}{\mathrm{3}}} \:−\:\boldsymbol{\mathrm{I}}\:\:=\:\:\boldsymbol{\mathrm{I}} \\ $$$$\:\:\:\:\:\:\:\frac{\boldsymbol{\pi}}{\mathrm{6}}\:\:=\:\:\mathrm{2}\boldsymbol{{I}}\:\:\:\:\:\:\:\boldsymbol{\mathrm{I}}=\:\frac{\boldsymbol{\pi}}{\mathrm{12}} \\ $$$$\:\boldsymbol{{ABDULAZIZ}}\:\:\:\boldsymbol{{ABDUVALIYEV}} \\ $$$$\: \\ $$