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

May 4, 2018

$- \frac{1}{9} \left({x}^{2} - 2\right) \sqrt{{x}^{2} + 1} + C$.

#### Explanation:

We have, $\int - {x}^{3} / \sqrt{9 + 9 {x}^{2}} \mathrm{dx} = - \frac{1}{3} \int {x}^{3} / \sqrt{1 + {x}^{2}} \mathrm{dx}$.

Let, $I = \int {x}^{3} / \sqrt{1 + {x}^{2}} \mathrm{dx}$.

Subst. $x = \tan y , \text{ so that, } \mathrm{dx} = {\sec}^{2} y \mathrm{dy}$.

$\therefore I = \int {\tan}^{3} \frac{y}{\sqrt{1 + {\tan}^{2} y}} {\sec}^{2} y \mathrm{dy}$,

$= \int {\tan}^{3} \frac{y}{\sec} y {\sec}^{2} y \mathrm{dy}$,

$= \int {\tan}^{3} y \sec y \mathrm{dy}$,

$= \int {\tan}^{2} y \cdot \tan y \sec y \mathrm{dy}$,

$= \int \left({\sec}^{2} y - 1\right) \tan y \sec y \mathrm{dy}$.

Next we subst. $\sec y = t . \therefore \sec y \tan y \mathrm{dy} = \mathrm{dt}$.

$\therefore I = \int \left({t}^{2} - 1\right) \mathrm{dt}$,

$= {t}^{3} / 3 - t$,

$= {\sec}^{3} \frac{y}{3} - \sec y \ldots \ldots \ldots \ldots \left[\because , t = \sec y\right]$,

$= \frac{1}{3} \sec y \left({\sec}^{2} y - 3\right)$,

$= \frac{1}{3} \sqrt{{\sec}^{2} y} \left\{\left(1 + {\tan}^{2} y\right) - 3\right\}$.

$= \frac{1}{3} \sqrt{1 + {\tan}^{2} y} \left({\tan}^{2} y - 2\right)$.

$\Rightarrow I = \frac{1}{3} \left({x}^{2} - 2\right) \sqrt{{x}^{2} + 1} \ldots \ldots \left[\because , \tan y = x\right]$.

Hence, $\int - {x}^{3} / \sqrt{9 + 9 {x}^{2}} \mathrm{dx} = - \frac{1}{9} \left({x}^{2} - 2\right) \sqrt{{x}^{2} + 1} + C$.

May 4, 2018

$- \frac{1}{9} \left({x}^{2} - 2\right) \sqrt{{x}^{2} + 1} + C$.

#### Explanation:

I have solved the Problem using trigo. substn., but the same

can be done without it, as shown below :

$I = \int - {x}^{3} / \sqrt{9 + 9 {x}^{2}} \mathrm{dx}$,

$= - \frac{1}{3} \int \frac{{x}^{2} \cdot x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}$,

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

$= - \frac{1}{3} \int \left\{\frac{{x}^{2} + 1}{\sqrt{{x}^{2} + 1}} - \frac{1}{\sqrt{{x}^{2} + 1}}\right\} x \mathrm{dx}$,

$= - \frac{1}{3} \int \left\{\sqrt{{x}^{2} + 1} - \frac{1}{\sqrt{{x}^{2} + 1}}\right\} x \mathrm{dx}$,

$= - \frac{1}{3} \cdot \frac{1}{2} \int \left\{\sqrt{{x}^{2} + 1} - \frac{1}{\sqrt{{x}^{2} + 1}}\right\} \left(2 x\right) \mathrm{dx}$,

$= - \frac{1}{6} \int \left\{{\left({x}^{2} + 1\right)}^{\frac{1}{2}} - {\left({x}^{2} + 1\right)}^{- \frac{1}{2}}\right\} d \left({x}^{2} + 1\right)$,

$= - \frac{1}{6} \left\{{\left({x}^{2} + 1\right)}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right) - {\left({x}^{2} + 1\right)}^{- \frac{1}{2} + 1} / \left(- \frac{1}{2} + 1\right)\right\}$,

$= - \frac{1}{6} \left\{\frac{2}{3} {\left({x}^{2} + 1\right)}^{\frac{3}{2}} - 2 {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right\}$,

$= - \frac{1}{6} \cdot \frac{2}{3} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} \left\{{\left({x}^{2} + 1\right)}^{\frac{3}{2} - \frac{1}{2}} - 3 {\left({x}^{2} + 1\right)}^{\frac{1}{2} - \frac{1}{2}}\right\}$,

$= - \frac{1}{9} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} \left\{\left({x}^{2} + 1\right) - 3\right\}$,

$= - \frac{1}{9} \left({x}^{2} - 2\right) \sqrt{{x}^{2} + 1} + C$, as before!