Oct 3, 2005 #1 trap Messages 51 Reaction score 0 Anyone know how to solve this? \text 8\pi\int_{0}^{\infty}\frac{t^3}{(4+t^2)^\frac{5}{2}} dt
Oct 3, 2005 #2 hotvette Homework Helper Messages 1,001 Reaction score 11 Sure, make the following substitution: u = 4 + t^2[/itex]
Oct 3, 2005 #3 trap Messages 51 Reaction score 0 yeah, i tried that, but i can't go further coz i can't get rid of the t^3
Oct 3, 2005 #4 hotvette Homework Helper Messages 1,001 Reaction score 11 No need to get rid of it. Take it further. I believe you'll end up with 2 integrals in fractional powers of u. Last edited: Oct 4, 2005
No need to get rid of it. Take it further. I believe you'll end up with 2 integrals in fractional powers of u.
Oct 3, 2005 #5 trap Messages 51 Reaction score 0 can you help me with one more step? i really can't make two integrals with power of u..
Oct 4, 2005 #7 dextercioby Science Advisor Insights Author Messages 13,394 Reaction score 4,064 HINT: \sqrt{4+t^{2}}=u t \ dt= u \ du \int_{0}^{\infty} \frac{t^{3}+4t-4t}{\left(\sqrt{t^{2}+4}\right)^{5}} \ dt =\int_{0}^{\infty} \frac{t^{2}+4}{\left(\sqrt{t^{2}+4}\right)^{5}}t \ dt - 4\int_{0}^{\infty}\frac{t \ dt}{\left(\sqrt{t^{2}+4}\right)^{5}} Daniel.
HINT: \sqrt{4+t^{2}}=u t \ dt= u \ du \int_{0}^{\infty} \frac{t^{3}+4t-4t}{\left(\sqrt{t^{2}+4}\right)^{5}} \ dt =\int_{0}^{\infty} \frac{t^{2}+4}{\left(\sqrt{t^{2}+4}\right)^{5}}t \ dt - 4\int_{0}^{\infty}\frac{t \ dt}{\left(\sqrt{t^{2}+4}\right)^{5}} Daniel.
Oct 5, 2005 #9 benorin Science Advisor Insights Author Messages 1,442 Reaction score 191 Let t^2=x-4 -> 2tdt=dx and 0<=t<=infinity -> 4<=x<=infinity so the integral becomes 4*Pi*Int((x-4)*x^(-5/2),x=4..infinity) = 8*Pi/3 excuse my maple notation, I have yet to learn LaTeX.
Let t^2=x-4 -> 2tdt=dx and 0<=t<=infinity -> 4<=x<=infinity so the integral becomes 4*Pi*Int((x-4)*x^(-5/2),x=4..infinity) = 8*Pi/3 excuse my maple notation, I have yet to learn LaTeX.
Oct 5, 2005 #10 saltydog Science Advisor Homework Helper Messages 1,590 Reaction score 3 Nice. Here we go: (just do a click on the equation and a pop-up window will display the LaTex commands) <br /> \begin{align*}<br /> 8\pi\int_0^{\infty}\frac{t^3}{(4+t^2)^{5/2}}dt &=<br /> \frac{8\pi}{2}\int_4^{\infty}\frac{x-4}{x^{5/2}}dx \\ &=<br /> 4\pi\int_4^{\infty}x^{-5/2}(x-4)dx \\ &=<br /> 4\pi\int_4^{\infty}(x^{-3/2}-4x^{-5/2})dx \\ &=<br /> 4\pi\left(8/3x^{-5/2}-2x^{-1/2}\right)_4^{\infty} \\ &=<br /> 4\pi\left(-(1/3-1)\right) \\ &=<br /> \frac{8\pi}{3}<br /> \end{align}<br />
Nice. Here we go: (just do a click on the equation and a pop-up window will display the LaTex commands) <br /> \begin{align*}<br /> 8\pi\int_0^{\infty}\frac{t^3}{(4+t^2)^{5/2}}dt &=<br /> \frac{8\pi}{2}\int_4^{\infty}\frac{x-4}{x^{5/2}}dx \\ &=<br /> 4\pi\int_4^{\infty}x^{-5/2}(x-4)dx \\ &=<br /> 4\pi\int_4^{\infty}(x^{-3/2}-4x^{-5/2})dx \\ &=<br /> 4\pi\left(8/3x^{-5/2}-2x^{-1/2}\right)_4^{\infty} \\ &=<br /> 4\pi\left(-(1/3-1)\right) \\ &=<br /> \frac{8\pi}{3}<br /> \end{align}<br />