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Question Number 200778 by sonukgindia last updated on 23/Nov/23

Answered by MM42 last updated on 23/Nov/23

$$ifn=1\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{1}{1+x^2}dx$$$${={tan}^{-1}\left(x\right)]}_0^{\mathrm{\infty }}=\frac{\pi }{2}$$$$forn>1$$$$x=tanu\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{1+{tan}^{2}u}{{(1+{tan}^{2}u)}^n}du={\int }_0^{\frac{\pi }{2}}{cos}^{2n-2}xdx=I_{2n-2}$$$$by‘‘expectforexpect{}^{″}$$$${\Rightarrow I_{2n-2}=sinu\times {cos}^{2n-3}u]}_0^{\frac{\pi }{2}}+(2n-3){\int }_0^{\frac{\pi }{2}}{cos}^{2n-4}u\times (1-{cos}^{2}u)du$$$$\Rightarrow I_{2n-2}=\frac{2n-3}{2n-2}{I}_{2n-4}=\frac{2n-3}{2n-2}\times \frac{2n-5}{2n-4}\times ...\times \frac{1}{2}{I}_2$$$$=\frac{2n-3}{2n-2}\times \frac{2n-5}{2n-4}\times ...\times \frac{1}{2}\times \frac{\pi }{2}✓$$$$$$

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