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Question Number 96280 by bobhans last updated on 31/May/20

Commented by bemath last updated on 31/May/20

$$\mathrm{let}\mathrm{w}=\arctan \mathrm{x}$$$$\mathrm{dw}=\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}\Rightarrow \int \mathrm{w}\mathrm{dw}={\left[\frac{1}{2}{\mathrm{w}}^2\right]}_0^{\frac{\pi }{2}}$$$$=\frac{1}{2}{\left(\frac{\pi }{2}\right)}^2=\frac{{\pi }^2}{8}.$$

Answered by 675480065 last updated on 31/May/20

$$\mathrm{let}\mathrm{u}=\mathrm{arctanx}$$$$\Rightarrow \mathrm{tanu}=\mathrm{x}$$$$\Rightarrow {\sec }^2\mathrm{udu}=\mathrm{dx}$$$$\Rightarrow (1+{\tan }^2\mathrm{u})\mathrm{du}=\mathrm{dx}\{\mathrm{but}\mathrm{tanu}=\mathrm{x}\}$$$$\Rightarrow (1+{\mathrm{x}}^2)\mathrm{du}=\mathrm{dx}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\left(\frac{\mathrm{arctanx}}{{\mathrm{x}}^2+1}\right)\mathrm{dx}={\int }_0^{\mathrm{\infty }}\left(\frac{({\mathrm{x}}^2+1)\mathrm{udu}}{{\mathrm{x}}^2+1}\right)$$$$={\int }_0^{\mathrm{\infty }}\mathrm{udu}$$$${\left[\frac{{\tan }^{-2}\mathrm{x}}{2}\right]}_0^{\mathrm{\infty }}=\frac{{\pi }^2}{8}$$

Commented by bobhans last updated on 31/May/20

$$\mathrm{good}.\mathrm{thank}\mathrm{you}$$

Answered by mathmax by abdo last updated on 31/May/20

$$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{\arctan \left(\mathrm{x}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}\mathrm{by}\mathrm{parts}{\mathrm{u}}^{{}^{\prime }}=\frac{1}{{\mathrm{x}}^2+1}\mathrm{and}\mathrm{v}=\mathrm{arctanx}$$$$\mathrm{A}={\left[{\arctan }^2\mathrm{x}\right]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{arctanx}}{1+{\mathrm{x}}^2}\mathrm{dx}=\frac{{\pi }^2}{4}-\mathrm{A}\Rightarrow 2\mathrm{A}=\frac{{\pi }^2}{4}\Rightarrow \mathrm{A}=\frac{{\pi }^2}{8}$$

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