Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 54585 by peter frank last updated on 07/Feb/19

$$showthat{\int }_0^{\mathrm{\infty }}\frac{x}{1+x^6}dx=\frac{\pi }{3\sqrt{3}}$$$$$$

Commented by maxmathsup by imad last updated on 07/Feb/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{x}{1+x^6}dxchangement{x}^6=tgivex=t^{\frac{1}{6}}\Rightarrow$$$$I={\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{6}}}{1+t}\frac{1}{6}{t}^{\frac{1}{6}-1}dt=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{3}-1}}{1+t}dtbutwehaveprovedthat$$$${\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}with0<a<1\Rightarrow I=\frac{1}{6}\frac{\pi }{sin\left(\frac{\pi }{3}\right)}=\frac{\pi }{6\frac{\sqrt{3}}{2}}=\frac{\pi }{3\sqrt{3}}.$$

Answered by arvinddayama00@gmail.com last updated on 08/Feb/19

$$x^2=tx=0,t=0$$$$2xdx=dtx=\mathrm{\infty },t=\mathrm{\infty }$$$$\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^3}$$$$\because {\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^{\alpha }}=\frac{\pi }{\alpha sin\left(\frac{\pi }{\alpha }\right)}★★$$$$so\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^3}=\frac{1}{2}.\frac{\pi }{3.\frac{\sqrt{3}}{2}}=\frac{\pi }{3\sqrt{3}}$$$$pls..cheak.....$$$$$$

Commented by peter frank last updated on 07/Feb/19

$$rightsir$$

Commented by rahul 19 last updated on 08/Feb/19

$$Sir,how{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^{\alpha }}=\frac{\pi }{\alpha \sin \left(\frac{\pi }{\alpha }\right)}?$$

Commented by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19

$$x^{\alpha }={tan}^2\theta x={\left(tan\theta \right)}^{\frac{2}{\alpha }}$$$$dx=\frac{2}{\alpha }{\left(tan\theta \right)}^{\frac{2}{\alpha }-1}\left({sec}^2\theta \right)d\theta$$$${\int }_0^{\frac{\pi }{2}}\frac{2}{\alpha }{\left(tan\theta \right)}^{\frac{2}{\alpha }-1}d\theta$$$$\frac{2}{\alpha }{\int }_0^{\frac{\pi }{2}}{\left(sin\theta \right)}^{\frac{2}{\alpha }-1}{\left(cos\theta \right)}^{1-\frac{2}{\alpha }}d\theta$$$$nowgammabetafunctionformula$$$$2{\int }_0^{\frac{\pi }{2}}{\left(sin\theta \right)}^{2p-1}{\left(cos\theta \right)}^{2q-1}d\theta =\frac{\lceil \left(p\right)\lceil q)}{\lceil (p+q)}$$$$nowhere2p-1=\frac{2}{\alpha }-1sop=\frac{1}{\alpha }$$$$2q-1=1-\frac{2}{\alpha }soq=(1-\frac{1}{\alpha })$$$$nowansweris=\frac{1}{\alpha }\times \frac{\lceil \left(\frac{1}{\alpha }\right)\lceil (1-\frac{1}{\alpha })}{\lceil \left(1\right)}=\frac{1}{\alpha }\times \frac{\pi }{sin\left(\frac{\pi }{\alpha }\right)}$$$$[againformula\lceil \left(a\right)\lceil (1-a)=\frac{\pi }{sin\left(a\pi \right)}]$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com