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

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

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

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

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

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

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