How do you integrate #int x^3/(sqrt(x^2-4))dx# using trigonometric substitution?

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Answer 1 · 2018-06-13T17:17:49

The answer is #=1/3(x^2-4)^(3/2)+4sqrt(x^2-4)+C#

There is no need for trigonometric substitution.

Let #u=x^2-4#, #=>#, #du=2xdx#

Therefore, the integral is

#I=int(x^3dx)/sqrt(x^2-4)=1/2int((u+4)du)/sqrtu#

#=1/2int(u^(1/2)+4u^(-1/2))du#

#=1/2(u^(3/2)/(3/2)+(4u^(1/2)/(1/2)))#

#=1/3u^(3/2)+4u^(1/2)#

#=1/3(x^2-4)^(3/2)+4sqrt(x^2-4)+C#