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Question Number 182045 by Emrice last updated on 03/Dec/22

$$$$$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1}{1+{\left(tanx\right)}^{2013}}dx=?$$$$$$

Commented by Frix last updated on 03/Dec/22

$$f\left(x\right)=\frac{1}{1+{\tan }^{n}x}\mathrm{is}\mathrm{symmetric}\mathrm{to}\mathrm{the}\mathrm{point}(\frac{\pi }{4};\frac{1}{2})\Rightarrow$$$$\underset{\frac{\pi }{4}-a}{\stackrel{\frac{\pi }{4}+a}{\int }}\frac{dx}{1+{\tan }^{n}x}=a\mathrm{\forall }n\in \mathbb{N}\Rightarrow$$$$\mathrm{answer}\mathrm{is}\frac{\pi }{12}$$

Answered by Ar Brandon last updated on 03/Dec/22

$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1}{1+{\left(\tan x\right)}^{2013}}dx$$$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\left(\cos x\right)}^{2013}}{{\left(\cos x\right)}^{2013}+{\left(\sin x\right)}^{2013}}dx...\mathrm{eqn}\left(i\right)$$$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\left(\sin x\right)}^{2013}}{{\left(\cos x\right)}^{2013}+{\left(\sin x\right)}^{2013}}dx...\mathrm{eqn}\left(ii\right)$$$$\mathrm{eqn}\left(i\right)+\mathrm{eqn}\left(ii\right)$$$$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\left(\cos x\right)}^{2013}+{\left(\sin x\right)}^{2013}}{{\left(\cos x\right)}^{2013}+{\left(\sin x\right)}^{2013}}dx={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx$$$$\Rightarrow I=\frac{1}{2}{\left[x\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{1}{2}(\frac{\pi }{3}-\frac{\pi }{6})=\frac{1}{2}\left(\frac{\pi }{6}\right)=\overline{)\begin{array}{c}\frac{\pi }{12}\end{array}}$$

Answered by SEKRET last updated on 04/Dec/22

$${\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}\left(\mathbf{x}\right)\mathbf{dx}={\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}(\mathbf{a}+\mathbf{b}-\mathbf{x})\mathbf{dx}$$$${\int }_{\frac{\mathit{\pi }}{6}}^{\frac{\mathit{\pi }}{3}}\frac{1}{1+{\left(\mathbf{tanx}\right)}^{2013}}\mathbf{dx}=\mathbf{I}$$$${\int }_{\frac{\mathit{\pi }}{6}}^{\frac{\mathit{\pi }}{3}}\frac{1}{1+{\left(\mathbf{tan}(\frac{\mathit{\pi }}{3}+\frac{\mathit{\pi }}{6}-x)\right)}^{2013}}\mathbf{dx}=$$$$={\int }_{\frac{\mathit{\pi }}{6}}^{\frac{\mathit{\pi }}{3}}\frac{1}{1+\frac{1}{{\left(\mathbf{tanx}\right)}^{2013}}}\mathbf{dx}={\int }_{\frac{\mathit{\pi }}{6}}^{\frac{\pi }{3}}\frac{{\left(\mathbf{tanx}\right)}^{2013}+1-1}{1+{\left(\mathbf{tanx}\right)}^{2013}}\mathbf{dx}=$$$${\int }_{\frac{\pi }{6}}^{\frac{\mathit{\pi }}{3}}1\mathbf{dx}-{\int }_{\frac{\mathit{\pi }}{6}}^{\frac{\mathit{\pi }}{3}}\frac{1}{1+{\left(\mathbf{tanx}\right)}^{2013}}\mathbf{dx}=\mathbf{I}$$$$\mathbf{x}{/}_{\frac{\mathit{\pi }}{6}}^{\frac{\mathit{\pi }}{3}}-\mathbf{I}=\mathbf{I}$$$$\frac{\mathit{\pi }}{6}=2\mathit{I}\mathbf{I}=\frac{\mathit{\pi }}{12}$$$$\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{L}\mathit{A}\mathit{Z}\mathit{I}\mathit{Z}\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{V}\mathit{A}\mathit{L}\mathit{I}\mathit{Y}\mathit{E}\mathit{V}$$$$$$

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