# How do you evaluate \int x ^ { 4} \sqrt { x ^ { 2} - 4} d x?

Dec 3, 2017

=$\frac{1}{6} \cdot \left({x}^{5} - {x}^{3} - 6 x\right) \cdot \sqrt{{x}^{2} - 4}$-$4 \ln \left(x + \sqrt{{x}^{2} - 4}\right) + C$

#### Explanation:

$\int {x}^{4} \cdot \sqrt{{x}^{2} - 4} \cdot \mathrm{dx}$

After using $x = 2 \sec u$ and $\mathrm{dx} = 2 \sec u \cdot \tan u \cdot \mathrm{du}$ transforms, this integral became,

$\int {\left(2 \sec u\right)}^{4} \cdot \sqrt{{\left(2 \sec u\right)}^{2} - 4} \cdot 2 \sec u \cdot \tan u \cdot \mathrm{du}$

=$\int 32 {\left(\sec u\right)}^{5} \cdot \tan u \cdot \sqrt{4 {\left(\tan u\right)}^{2}} \cdot \mathrm{du}$

=$\int 64 {\left(\sec u\right)}^{5} \cdot {\left(\tan u\right)}^{2} \cdot \mathrm{du}$

=$\int 64 {\left(\sec u\right)}^{5} \cdot \left[{\left(\sec u\right)}^{2} - 1\right] \cdot \mathrm{du}$

=$\int 64 {\left(\sec u\right)}^{7} \cdot \mathrm{du}$-$\int 64 {\left(\sec u\right)}^{5} \cdot \mathrm{du}$

After using $\int {\left(\sec u\right)}^{n} \cdot \mathrm{du}$=$\frac{1}{n - 1} \cdot {\left(\sec u\right)}^{n - 2} \cdot \tan u$+$\frac{n - 2}{n - 1} \cdot \int {\left(\sec u\right)}^{n} \cdot \mathrm{du}$ reduction formula,

=$\int 64 {\left(\sec u\right)}^{7} \cdot \mathrm{du}$-$\int 64 {\left(\sec u\right)}^{5} \cdot \mathrm{du}$

=64*[1/6*(secu)^5*tanu+5/6*int (secu)^5*du]-$\int 64 {\left(\sec u\right)}^{5} \cdot \mathrm{du}$

=$\frac{32}{3} \cdot {\left(\sec u\right)}^{5} \cdot \tan u + \frac{160}{3} \cdot \int {\left(\sec u\right)}^{5} \cdot \mathrm{du}$-$\int 64 {\left(\sec u\right)}^{5} \cdot \mathrm{du}$

=$\frac{32}{3} \cdot {\left(\sec u\right)}^{5} \cdot \tan u - \frac{32}{3} \cdot \int {\left(\sec u\right)}^{5} \cdot \mathrm{du}$

=$\frac{32}{3} \cdot {\left(\sec u\right)}^{5} \cdot \tan u$-32/3*[1/4*(secu)^3*tanu+$\frac{3}{4} \cdot \int {\left(\sec u\right)}^{3} \cdot \mathrm{du}$]

=$\frac{32}{3} \cdot {\left(\sec u\right)}^{5} \cdot \tan u$-$\frac{8}{3} \cdot {\left(\sec u\right)}^{3} \cdot \tan u$-$8 \int {\left(\sec u\right)}^{3} \cdot \mathrm{du}$

=$\frac{32}{3} \cdot {\left(\sec u\right)}^{5} \cdot \tan u$-$\frac{8}{3} \cdot {\left(\sec u\right)}^{3} \cdot \tan u$-8*[1/2*secu*tanu+$\frac{1}{2} \cdot \int \sec u \cdot \mathrm{du}$]

=$\frac{32}{3} \cdot {\left(\sec u\right)}^{5} \cdot \tan u$-$\frac{8}{3} \cdot {\left(\sec u\right)}^{3} \cdot \tan u$-$4 \sec u \cdot \tan u$-$4 \int \sec u \cdot \mathrm{du}$

=$\frac{32}{3} \cdot {\left(\sec u\right)}^{5} \cdot \tan u$-$\frac{8}{3} \cdot {\left(\sec u\right)}^{3} \cdot \tan u$-$4 \sec u \cdot \tan u$-$4 \ln \left(\sec u + \tan u\right) + C 1$

After using $x = 2 \sec u$, $\sec u = \frac{x}{2}$ and $\tan u = \frac{\sqrt{{x}^{2} - 4}}{2}$ inverse transformations, I found

$\int {x}^{4} \cdot \sqrt{{x}^{2} - 4} \cdot \mathrm{dx}$

=$\left(\frac{1}{6} \cdot {x}^{5} - \frac{1}{6} \cdot {x}^{3} - x\right) \cdot \sqrt{{x}^{2} - 4}$-$4 \ln \left[\frac{x + \sqrt{{x}^{2} - 4}}{2}\right] + C 1$

=$\frac{1}{6} \cdot \left({x}^{5} - {x}^{3} - 6 x\right) \cdot \sqrt{{x}^{2} - 4}$-$4 \ln \left(x + \sqrt{{x}^{2} - 4}\right) + C$

Note: $C = C 1 + 4 L n 2$