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Question Number 209217 by mnjuly1970 last updated on 04/Jul/24

$$$$$$\mathit{c}\mathit{a}\mathit{l}\mathit{c}\mathit{u}\mathit{l}\mathit{a}\mathit{t}\mathit{e}:$$$$$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{tan}^{-1}\left(x\right)}{{(1+x^2)}^2}dx=?$$$$$$

Answered by Berbere last updated on 04/Jul/24

$${\tan }^{-1}\left(x\right)=u$$$${\int }_0^{\frac{\pi }{2}}{ucos}^2\left(u\right)du={\int }_0^{\frac{\pi }{2}}u.\left(\frac{1+cos\left(2u\right)}{2}\right)$$$$=\frac{1}{4}.\frac{{\pi }^2}{4}+\frac{1}{2}{\left[\frac{sin\left(2u\right)}{2}u\right]}_0^{\frac{\pi }{2}}-\frac{1}{4}\int sin\left(2u\right)du$$$$=\frac{{\pi }^2}{16}+\frac{1}{8}{\left[cos\left(2u\right)\right]}_0^{\frac{\pi }{2}}=\frac{{\pi }^2}{16}-\frac{1}{4}$$

Answered by lepuissantcedricjunior last updated on 05/Jul/24

$$\mathit{i}={\int }_0^{\mathrm{\infty }}\frac{\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)}{{(1+{\mathit{x}}^2)}^2}\mathit{d}\mathit{x}$$$$\mathit{p}\mathit{o}\mathit{s}\mathit{o}\mathit{n}\mathit{s}\mathit{x}=\mathit{t}\mathit{a}\mathit{n}\mathit{t}=>\mathit{d}\mathit{x}=(1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{t})\mathit{d}\mathit{t}$$$$\mathit{i}={\int }_0^{\mathit{\pi }/2}\frac{\mathit{t}}{1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{t}}\mathit{d}\mathit{t}={\int }_0^{\frac{\pi }{2}}{\mathit{t}\mathit{c}\mathit{o}\mathit{s}}^2\left(\mathit{t}\right)\mathit{d}\mathit{t}$$$$\{\begin{array}{l}\mathit{u}=\mathit{t}\\ {\mathit{v}}^{\prime }={\mathit{c}\mathit{o}\mathit{s}}^2\left(\mathit{t}\right)\end{array}=>\{\begin{array}{l}{\mathit{u}}^{\prime }=1\\ \mathit{v}=\frac{\mathit{t}}{2}+\frac{\mathit{s}\mathit{i}\mathit{n}2\mathit{t}}{4}\end{array}$$$$\mathit{i}={[\frac{{\mathit{t}}^2}{2}+\frac{\mathit{t}\mathit{s}\mathit{i}\mathit{n}2\mathit{t}}{4}]}_0^{\frac{\mathit{\pi }}{2}}-{\int }_0^{\frac{\mathit{\pi }}{2}}(\frac{\mathit{t}}{2}+\frac{\mathit{s}\mathit{i}\mathit{n}2\mathit{t}}{4})\mathit{d}\mathit{t}$$$$=\frac{{\mathit{\pi }}^2}{8}-{[\frac{{\mathit{t}}^2}{4}-\frac{\mathit{c}\mathit{o}\mathit{s}2\mathit{t}}{8}]}_0^{\frac{\mathit{\pi }}{2}}$$$$=\frac{{\mathit{\pi }}^2}{8}-\frac{{\mathit{\pi }}^2}{16}-\frac{1}{8}-\frac{1}{8}=\frac{{\mathit{\pi }}^2}{16}-\frac{1}{4}$$$$\Rightarrow \mathit{i}={\int }_0^{\mathrm{\infty }}\frac{\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)}{{(1+{\mathit{x}}^2)}^2}\mathit{d}\mathit{x}=\frac{{\mathit{\pi }}^2}{16}-\frac{1}{4}$$

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