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

Aug 14, 2016

$- \frac{2}{35} {\left(3 - x\right)}^{\frac{3}{2}} \left(24 + 12 x + 5 {x}^{2}\right) + C$.

#### Explanation:

To get rid of the Radical Sign, i.e., the Square Root, let us subst.

$3 - x = {t}^{2} , \mathmr{and} , x = 3 - {t}^{2} , s o , \mathrm{dx} = - 2 t \mathrm{dt}$. Hence,

$I = \int {x}^{2} \sqrt{3 - x} \mathrm{dx} = \int {\left(3 - {t}^{2}\right)}^{2} \cdot t \cdot \left(- 2 t\right) \mathrm{dt}$

$= - 2 \int \left(9 - 6 {t}^{2} + {t}^{4}\right) {t}^{2} \mathrm{dt}$

$= - 2 \int \left(9 {t}^{2} - 6 {t}^{4} + {t}^{6}\right) \mathrm{dt}$

$= - 18 \int {t}^{2} \mathrm{dt} + 12 \int {t}^{4} \mathrm{dt} - 2 \int {t}^{6} \mathrm{dt}$

$= - 18 \cdot {t}^{3} / 3 + 12 \cdot {t}^{5} / 5 - 2 \cdot {t}^{7} / 7$

$= - 6 {t}^{3} + \frac{12}{5} {t}^{5} - \frac{2}{7} {t}^{7}$

$= - \frac{2}{35} \cdot {t}^{3} \left(105 - 42 {t}^{2} + 5 {t}^{4}\right)$

$= - \frac{2}{35} \cdot {\left({t}^{2}\right)}^{\frac{3}{2}} \left(105 - 42 {t}^{2} + 5 {t}^{4}\right)$

$= - \frac{2}{35} {\left(3 - x\right)}^{\frac{3}{2}} \left\{105 - 42 \left(3 - x\right) + 5 {\left(3 - x\right)}^{2}\right\}$

$= - \frac{2}{35} {\left(3 - x\right)}^{\frac{3}{2}} \left(24 + 12 x + 5 {x}^{2}\right) + C$.

Enjoy Maths.!