How do you integrate $int x^3sqrt(x^2-1)$ using substitution?

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Answer 1 · 2016-07-29T13:45:48

$1/15(3x^2+2)(x^2-1)^(3/2)+C$ .

Let us take the subst. $x^2-1=t^2$ , so, $x^2=t^2+1$ , &,

$2xdx=2tdt, or, xdx=tdt$

Now, $I=intx^3sqrt(x^2-1)dx=intx^2sqrt(x^2-1)xdx$

$=int(t^2+1)(sqrt(t^2))tdt=int(t^4+t^2)dt=t^5/5+t^3/3$

$=t^3/15(3t^2+5)=t/15*t^2(3t^2+5)$

$=sqrt(x^2-1)/15*(x^2-1){3(x^2-1)+5}$

$=1/15(3x^2+2)(x^2-1)^(3/2)+C$ .

Answer 2 · 2016-07-29T15:47:37

Here is a third solution.

$intx^3sqrt(x^2-1) dx = int x^2sqrt(x^2-1) x dx$

Let $u = x^2-1$ ,

so that $du = 2xdx$

and $x^2 = u+1$ .

The integral becomes

$int (u+1)u^(1/2) 1/2 du = 1/2 int (u^(3/2)+u^(1/2))du$

$= 1/2[2/5u^(5/2)+2/3u^(1/2)] +C$

$= 1/15[3u^(5/2)+5u^(3/2)]+C$

$= 1/15 u^(3/2)[3u+5]+C$

Back-substituting gets us

$= 1/15 (x^2-1)^(3/2)[3(x^2-1)+5]+C$

$= 1/15 (x^2-1)^(3/2)(3x^2+2)+C$

How do you integrate int x^3sqrt(x^2-1)int x^3sqrt(x^2-1) using substitution?