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Question Number 60960 by maxmathsup by imad last updated on 27/May/19

$$find{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}tan\left(\frac{1}{1+x^2}\right)dx$$

Commented by mathsolverby Abdo last updated on 28/May/19

$$letI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}tan\left(\frac{1}{1+x^2}\right)dx\Rightarrow$$$$I=2{\int }_0^{+\mathrm{\infty }}tan\left(\frac{1}{1+x^2}\right)dx{=}_{x=tan\theta }$$$$2{\int }_0^{\frac{\pi }{2}}tan\left({cos}^2\theta \right)(1+{tan}^2\theta )d\theta$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{tan\left({cos}^2\theta \right)}{{cos}^2\theta }d\theta ifwetake$$$$tan\left(u\right)\sim u+\frac{u^3}{3}weget$$$$tan\left({cos}^2\theta \right)\sim {cos}^2\theta +\frac{{cos}^6\theta }{3}\Rightarrow$$$$\frac{tan\left({cos}^2\theta \right)}{{cos}^2\theta }\sim 1+\frac{{cos}^4\theta }{3}\Rightarrow$$$$I\sim 2{\int }_0^{\frac{\pi }{2}}(1+\frac{{cos}^4\theta }{3})d\theta =\pi +\frac{2}{3}{\int }_0^{\frac{\pi }{2}}{cos}^4\theta d\theta$$$${\int }_0^{\frac{\pi }{2}}{cos}^4\theta d\theta =\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(1+cos{\left(2\theta \right)}^{2}d\theta$$$$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(1+2cos\theta +{cos}^2\left(2\theta \right))d\theta$$$$=\frac{\pi }{8}+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}cos\theta d\theta +\frac{1}{8}{\int }_0^{\frac{\pi }{2}}(1+cos\left(4\theta \right))d\theta$$$$=\frac{\pi }{8}+\frac{1}{2}+\frac{\pi }{16}+0$$$$=\frac{3\pi }{16}+\frac{1}{2}\Rightarrow I\sim \pi +\frac{2}{3}(\frac{3\pi }{16}+\frac{1}{2})\Rightarrow$$$$I\sim \pi +\frac{\pi }{8}+\frac{1}{3}\Rightarrow I\sim \frac{9\pi }{8}+\frac{1}{3}$$

Commented by mathsolverby Abdo last updated on 28/May/19

$$I\sim 3.86$$

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