# How do you integrate int (6x^3)/sqrt(9+x^2) dx using trigonometric substitution?

Jun 8, 2018

$2 \left({x}^{2} - 18\right) \sqrt{{x}^{2} + 9} + C$.

#### Explanation:

Suppose that, $I = \int \frac{6 {x}^{3}}{\sqrt{9 + {x}^{2}}} \mathrm{dx} = \int \frac{6 {x}^{3}}{\sqrt{{x}^{2} + {3}^{2}}} \mathrm{dx}$.

We subst. $x = 3 \tan u . \therefore \mathrm{dx} = 3 {\sec}^{2} u \mathrm{du}$.

$\therefore I = \int \frac{\left(6 \cdot 27 {\tan}^{3} u\right) \left(3 {\sec}^{2} u\right)}{\sqrt{9 {\tan}^{2} u + 9}} \mathrm{du}$,

$= 162 \int \left({\tan}^{3} u \sec u\right)$,

$162 \int \left({\tan}^{2} u \cdot \tan u \sec u\right) \mathrm{du}$.

$= 162 \int \left({\sec}^{2} u - 1\right) \sec u \tan u \mathrm{du}$.

If we take $\sec u = v , \text{ then, } \sec u \tan u \mathrm{du} = \mathrm{dv}$.

$\therefore I = 162 \int \left({v}^{2} - 1\right) \mathrm{dv}$,

$= 162 \left({v}^{3} / 3 - v\right)$,

$= \frac{162}{3} v \left({v}^{2} - 3\right)$,

$= 54 \sec u \left({\sec}^{2} u - 3\right) \ldots \ldots \ldots \ldots \left[\because , v = \sec u\right]$,

$= 54 \sqrt{{\tan}^{2} u + 1} \left\{\left({\tan}^{2} u + 1\right) - 3\right\}$,

$= 54 \left\{\sqrt{{\left(\frac{x}{3}\right)}^{2} + 1}\right\} \left\{{\left(\frac{x}{3}\right)}^{2} - 2\right\}$,

$\Rightarrow I = 2 \left({x}^{2} - 18\right) \sqrt{{x}^{2} + 9} + C$, as desired!

Feel the Joy of Maths.!

Jun 8, 2018

Here is a ${2}^{n d}$ Method ( without using trgo. substn.) :

$2 \left({x}^{2} - 18\right) \sqrt{{x}^{2} + 9} + C$

#### Explanation:

Let, $I = \int \frac{6 {x}^{3}}{\sqrt{{x}^{2} + 9}} \mathrm{dx} = \int \frac{\left(3 {x}^{2}\right) \left(2 x \mathrm{dx}\right)}{\sqrt{{x}^{2} + 9}}$.

Subst. ${x}^{2} + 9 = {t}^{2.} \therefore {x}^{2} = {t}^{2} - 9. \therefore 2 x \mathrm{dx} = 2 t \mathrm{dt}$.

$\therefore I = \int \frac{3 \left({t}^{2} - 9\right) \left(2 t \mathrm{dt}\right)}{\sqrt{{t}^{2}}}$,

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

$= 6 \left({t}^{3} / 3 - 9 t\right)$,

$= 2 t \left({t}^{2} - 27\right)$,

$= 2 \left\{\left({x}^{2} + 9\right) - 27\right\} \sqrt{{x}^{2} + 9}$.

$\Rightarrow I = 2 \left({x}^{2} - 18\right) \sqrt{{x}^{2} + 9} + C$, as in the ${1}^{s t}$ Method!