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abdo799
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in this video , the prof had to integrate x/(a^2+x^2)^3/2 , i know we usually do this using substitution , but in the video...he ignored the x and integrate like it was 1/(a^2+x^2)^3/2, how does that work?
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SteamKing said:Look carefully at the integrand x/(a^2+x^2)^(2/3). What is the derivative of (a^2+x^2)? Is it x times some constant perhaps? Can you rewrite the integrand as the product of two expressions, rather than the quotient?
BTW, your video requires a login to view, so we can't see it.
abdo799 said:the power on the brakets is 3/2
SteamKing said:The power on the brackets is immaterial. The principle remains.
The formula for integration of x/(a^2+x^2)^2/3 is ∫x/(a^2+x^2)^2/3 dx = -1/(3a^2 √(a^2+x^2)) + C, where C is the constant of integration.
The number 'a' represents the radius of the sphere in which the volume is being calculated. It is the distance from the center of the sphere to the edge of the sphere.
Yes, the formula can be used for any value of 'a'. However, if 'a' is negative, the result of the integration will also be negative.
The integration of x/(a^2+x^2)^2/3 is considered difficult because it involves the use of trigonometric substitutions and various integration techniques such as integration by parts or u-substitution. It also requires a deep understanding of the fundamental principles of calculus.
Yes, the integration of x/(a^2+x^2)^2/3 has real-life applications in physics, particularly in calculating the electric field intensity due to a spherical charge distribution. It is also used in calculating the volume of spheres and finding the center of mass of a sphere.