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

Jul 29, 2016

$\frac{1}{15} \left(3 {x}^{2} + 2\right) {\left({x}^{2} - 1\right)}^{\frac{3}{2}} + C$.

Explanation:

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

$2 x \mathrm{dx} = 2 t \mathrm{dt} , \mathmr{and} , x \mathrm{dx} = t \mathrm{dt}$

Now, $I = \int {x}^{3} \sqrt{{x}^{2} - 1} \mathrm{dx} = \int {x}^{2} \sqrt{{x}^{2} - 1} x \mathrm{dx}$

$= \int \left({t}^{2} + 1\right) \left(\sqrt{{t}^{2}}\right) t \mathrm{dt} = \int \left({t}^{4} + {t}^{2}\right) \mathrm{dt} = {t}^{5} / 5 + {t}^{3} / 3$

$= {t}^{3} / 15 \left(3 {t}^{2} + 5\right) = \frac{t}{15} \cdot {t}^{2} \left(3 {t}^{2} + 5\right)$

$= \frac{\sqrt{{x}^{2} - 1}}{15} \cdot \left({x}^{2} - 1\right) \left\{3 \left({x}^{2} - 1\right) + 5\right\}$

$= \frac{1}{15} \left(3 {x}^{2} + 2\right) {\left({x}^{2} - 1\right)}^{\frac{3}{2}} + C$.

Jul 29, 2016

Here is a third solution.

Explanation:

$\int {x}^{3} \sqrt{{x}^{2} - 1} \mathrm{dx} = \int {x}^{2} \sqrt{{x}^{2} - 1} x \mathrm{dx}$

Let $u = {x}^{2} - 1$,

so that $\mathrm{du} = 2 x \mathrm{dx}$

and ${x}^{2} = u + 1$.

The integral becomes

$\int \left(u + 1\right) {u}^{\frac{1}{2}} \frac{1}{2} \mathrm{du} = \frac{1}{2} \int \left({u}^{\frac{3}{2}} + {u}^{\frac{1}{2}}\right) \mathrm{du}$

$= \frac{1}{2} \left[\frac{2}{5} {u}^{\frac{5}{2}} + \frac{2}{3} {u}^{\frac{1}{2}}\right] + C$

$= \frac{1}{15} \left[3 {u}^{\frac{5}{2}} + 5 {u}^{\frac{3}{2}}\right] + C$

$= \frac{1}{15} {u}^{\frac{3}{2}} \left[3 u + 5\right] + C$

Back-substituting gets us

$= \frac{1}{15} {\left({x}^{2} - 1\right)}^{\frac{3}{2}} \left[3 \left({x}^{2} - 1\right) + 5\right] + C$

$= \frac{1}{15} {\left({x}^{2} - 1\right)}^{\frac{3}{2}} \left(3 {x}^{2} + 2\right) + C$