- #1
latentcorpse
- 1,444
- 0
I need to do the following integral:
[itex]B_z=\frac{\mu_0 I N a^2}{2} \int_{0}^{L} (a^2+(z-z')^2)^{-\frac{3}{2}} dz'[/itex]
so far i have set [itex]x=z-z' \Rightarrow dx=-dz'[/itex] and the limits also change to give
[itex]B_z = - \frac{\mu_0 I N a^2}{2} \int_{z}^{z-L} \frac{dx}{(a^2+x^2)^{\frac{3}{2}}}=\frac{\mu_0 I N a^2}{2a^3} \int_{z-L}^{z} \frac{dx}{(1+(\frac{x}{a})^2)^{\frac{3}{2}}}[/itex]
im having trouble with the power of 3/2 as otherwise its just a standard integral to arctan. any advice?
[itex]B_z=\frac{\mu_0 I N a^2}{2} \int_{0}^{L} (a^2+(z-z')^2)^{-\frac{3}{2}} dz'[/itex]
so far i have set [itex]x=z-z' \Rightarrow dx=-dz'[/itex] and the limits also change to give
[itex]B_z = - \frac{\mu_0 I N a^2}{2} \int_{z}^{z-L} \frac{dx}{(a^2+x^2)^{\frac{3}{2}}}=\frac{\mu_0 I N a^2}{2a^3} \int_{z-L}^{z} \frac{dx}{(1+(\frac{x}{a})^2)^{\frac{3}{2}}}[/itex]
im having trouble with the power of 3/2 as otherwise its just a standard integral to arctan. any advice?