How do you find the antiderivative of #int xsqrt(100-x^2)dx#?

1 Answer
mason m
Nov 6, 2016

#-(100-x^2)^(3/2)/3+C#

Explanation:

#I=intxsqrt(100-x^2)dx#

The best substitution to make here is #u=100-x^2#. This implies that #(du)/dx=-2x#, so #du=-2xdx#.

Since we have just #xdx# in the integral, multiply the integral's interior by #-2#. To balance this out, multiply the exterior of the integral by #-1//2#.

#I=-1/2int(-2x)sqrt(100-x^2)dx#

#I=-1/2intunderbrace(sqrt(100-x^2))_sqrtuoverbrace((-2xdx))^(du)#

#I=-1/2intsqrtudu#

This can be integrated if we write the square root using a fractional power.

#I=-1/2intu^(1/2)du#

This can be integrated using the power rule for integration, or #intu^ndu=u^(n+1)/(n+1)+C#. So:

#I=-1/2(u^(1/2+1)/(1/2+1))+C=-1/2(u^(3/2)/(3/2))+C#

#I=-1/2(2/3)u^(3/2)+C=-u^(3/2)/3+C#

Since #u=100-x^2#:

#I=-(100-x^2)^(3/2)/3+C#

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