# How do you integrate int x^2/sqrt(x^2-81)dx using trigonometric substitution?

May 31, 2018

$I = \frac{x}{2} \sqrt{{x}^{2} - 81} + \frac{81}{2} \ln | x + \sqrt{{x}^{2} - 81} | + C$

#### Explanation:

Here,

$I = \int {x}^{2} / \sqrt{{x}^{2} - 81} \mathrm{dx}$

Subst. $x = 9 \sec u \implies \mathrm{dx} = 9 \sec u \tan u \mathrm{du} , \mathmr{and} \sec u = \frac{x}{9}$

So,

$I = \int \frac{81 {\sec}^{2} u}{\sqrt{81 {\sec}^{2} u - 81}} \times 9 \sec u \tan u \mathrm{du}$

$= \int \frac{81 {\sec}^{2} u}{\cancel{9} \cancel{\tan} u} \times \cancel{9} \sec u \cancel{\tan} u \mathrm{du}$

$= 81 \int \sec u \cdot {\sec}^{2} u \mathrm{du}$

$= 81 \int \sqrt{1 + {\tan}^{2} u} \cdot {\sec}^{2} u \mathrm{du}$

Let, $\tan u = t \implies {\sec}^{2} u \mathrm{du} = \mathrm{dt}$

$\therefore I = 81 \int \sqrt{1 + {t}^{2}} \mathrm{dt}$

$= 81 \left[\frac{t}{2} \sqrt{1 + {t}^{2}} + {1}^{2} / 2 \ln | t + \sqrt{1 + {t}^{2}} |\right] + c$

Subst. back , $t = \tan u$

$I$=$81 \left[\tan \frac{u}{2} \sqrt{1 + {\tan}^{2} u} + \frac{1}{2} \ln | \tan u + \sqrt{1 + {\tan}^{2} u} |\right] + C '$

=$81 \left[\frac{\sqrt{{\sec}^{2} u - 1}}{2} \times \sec u + \frac{1}{2} \ln | \sqrt{{\sec}^{2} u - 1} + \sec u |\right] + C '$

Again subst. $\sec u = \frac{x}{9}$

I=81[sqrt((x^2)/81-1))/2xxx/9+1/2ln|sqrt((x^2/81)-1)+x/9|]+C'

$= 81 \left[\frac{\sqrt{{x}^{2} - 81}}{2 \times 9} \cdot \frac{x}{9} + \frac{1}{2} \ln | \frac{\sqrt{{x}^{2} - 81}}{9} + \frac{x}{9} |\right] + C '$

$= \frac{x}{2} \sqrt{{x}^{2} - 81} + \frac{81}{2} \ln | \frac{x + \sqrt{{x}^{2} - 81}}{9} | + C '$
$= \frac{x}{2} \sqrt{{x}^{2} - 81} + \frac{81}{2} \left[\ln | x + \sqrt{{x}^{2} - 81} | - \ln 9\right] + C '$

$= \frac{x}{2} \sqrt{{x}^{2} - 81} + \frac{81}{2} \ln | x + \sqrt{{x}^{2} - 81} | - \frac{81}{2} \ln 9 + C '$

$= \frac{x}{2} \sqrt{{x}^{2} - 81} + \frac{81}{2} \ln | x + \sqrt{{x}^{2} - 81} | + C$

where $C = C ' - \frac{81}{2} \ln 9$

Jun 1, 2018

$\frac{x}{2} \sqrt{{x}^{2} - 81} + \frac{81}{2} \ln | \left(x + \sqrt{{x}^{2} - 81}\right) | + C .$

#### Explanation:

Kindly refer to the Second Method which does not use the

Trigo. Subst. to solve the Problem.

Prerequisite :

$\int \sqrt{{x}^{2} - {a}^{2}} \mathrm{dx} = \frac{x}{2} \sqrt{{x}^{2} - {a}^{2}} - {a}^{2} / 2 \ln | \left(x + \sqrt{{x}^{2} - {a}^{2}}\right) | + {c}_{1}$,

&, $\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \ln | \left(x + \sqrt{{x}^{2} - {a}^{2}}\right) | + {c}_{2}$,

Suppose that, $I = \int {x}^{2} / \sqrt{{x}^{2} - 81} \mathrm{dx}$.

$\therefore I = \int \frac{\left({x}^{2} - 81\right) + 81}{\sqrt{{x}^{2} - 81}} \mathrm{dx}$,

$= \int \left\{\frac{{x}^{2} - 81}{\sqrt{{x}^{2} - 81}} + \frac{81}{\sqrt{{x}^{2} - 81}}\right\} \mathrm{dx}$,

$= \int \sqrt{{x}^{2} - 81} \mathrm{dx} + 81 \int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 81}}$,

=x/2sqrt(x^2-81)-81/2ln|(x+sqrt(x^2-81)|

+81ln|(x+sqrt(x^2-81)|.

$\therefore I = \frac{x}{2} \sqrt{{x}^{2} - 81} + \frac{81}{2} \ln | \left(x + \sqrt{{x}^{2} - 81}\right) | + C$, as

Respected Maganbhai P. has already derived!.

Enjoy Maths.!