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# How do you integrate int x^2/(sqrt(x^2-4))dx using trigonometric substitution?

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#### Explanation

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#### Explanation:

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1
Mar 22, 2018

$I = \frac{x}{2} \sqrt{{x}^{2} - 4} + 2 \ln | x + \sqrt{{x}^{2} - 4} | + c$
If we take $x = 2 \sec \theta$ leads to $I = \int {\sec}^{3} \theta d \left(\theta\right) = \int \sqrt{1 + {\tan}^{2} \theta} {\sec}^{2} \theta d \left(\theta\right)$....Again take $\tan \theta = t \implies I = \int \sqrt{1 - {t}^{2}} \mathrm{dt}$,then use (1)

#### Explanation:

We have,
color(red)((1)intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-(a^2)/2ln|x+sqrt(x^2-a^2)|
color(red)((2)int1/sqrt(x^2-a^2)dx=ln|x+sqrt(x^2-4)|+c
Hence,
$I = \int \frac{{x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = \int \frac{\left({x}^{2} - 4\right) + 4}{\sqrt{{x}^{2} - 4}} \mathrm{dx}$
$I = \int \frac{{x}^{2} - 4}{\sqrt{{x}^{2} - 4}} \mathrm{dx} + \int \frac{4}{\sqrt{{x}^{2} - 4}} \mathrm{dx}$
$= \int \sqrt{{x}^{2} - 4} \mathrm{dx} + 4 \int \frac{1}{\sqrt{{x}^{2} - 4}} \mathrm{dx}$
$= \int \sqrt{{x}^{2} - {2}^{2}} \mathrm{dx} + 4 \int \frac{1}{\sqrt{{x}^{2} - {2}^{2}}} \mathrm{dx}$
Using (1) and (2)
$I = \frac{x}{2} \sqrt{{x}^{2} - {2}^{2}} - \frac{{2}^{2}}{2} \ln | x + \sqrt{{x}^{2} - {2}^{2}} | + 4 \ln | x + \sqrt{{x}^{2} - {2}^{2}} |$
$= \frac{x}{2} \sqrt{{x}^{2} - 4} - 2 \ln | x + \sqrt{{x}^{2} - 4} | + 4 \ln | x + \sqrt{{x}^{2} - 4} | + c$
$I = \frac{x}{2} \sqrt{{x}^{2} - 4} + 2 \ln | x + \sqrt{{x}^{2} - 4} | + c$

For Trigonometric substitution proceed according to Hint given
above

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Mar 22, 2018

$\frac{x}{2} \sqrt{{x}^{2} - 4} + 2 \ln | \frac{x}{2} + \frac{\sqrt{{x}^{2} - 4}}{2} | + C , \mathmr{and} ,$

$\frac{x}{2} \sqrt{{x}^{2} - 4} + 2 \ln | x + \sqrt{{x}^{2} - 4} | + c , c = C - 2 \ln 2$.

#### Explanation:

Let, $I = \int {x}^{2} / \sqrt{{x}^{2} - 4} \mathrm{dx}$.

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

$\therefore I = \int \frac{4 {\sec}^{2} y}{\sqrt{4 {\sec}^{2} y - 4}} \cdot 2 \sec y \tan y \mathrm{dy}$,

$= \int \frac{4 {\sec}^{2} y}{2 \tan y} \cdot 2 \sec y \tan y \mathrm{dy}$,

$= 4 \int {\sec}^{3} y \mathrm{dy}$,

$= 4 \int \sec y \cdot {\sec}^{2} y \mathrm{dy}$,

$= 4 \int \sqrt{{\tan}^{2} y + 1} {\sec}^{2} y \mathrm{dy}$.

Next, we use the substn. $\tan y = t \therefore {\sec}^{2} y \mathrm{dy} = \mathrm{dt}$.

$\therefore I = 4 \int \sqrt{{t}^{2} + 1} \mathrm{dt}$,

$= 4 \left\{\frac{t}{2} \sqrt{{t}^{2} + 1} + \frac{1}{2} \ln | t + \sqrt{{t}^{2} + 1} |\right\}$,

$= 2 \left\{\tan y \sec y + \ln | \tan y + \sec y |\right\} \ldots \ldots \left[\because , t = \tan y\right]$.

Returning to $\sec y = \frac{x}{2} , \text{ so that, } \tan y = \sqrt{{x}^{2} / 4 - 1}$, we have,

$I = 2 \left\{\frac{x}{2} \cdot \frac{\sqrt{{x}^{2} - 4}}{2} + \ln | \frac{x}{2} + \frac{\sqrt{{x}^{2} - 4}}{2} |\right\}$,

$\Rightarrow I = \frac{x}{2} \sqrt{{x}^{2} - 4} + 2 \ln | \frac{x}{2} + \frac{\sqrt{{x}^{2} - 4}}{2} | + C , \mathmr{and} ,$

$I = \frac{x}{2} \sqrt{{x}^{2} - 4} + 2 \ln | x + \sqrt{{x}^{2} - 4} | + c , c = C - 2 \ln 2$.

However, the integral can easily be dealt with without

using the substn. as shown below :

$I = \int {x}^{2} / \sqrt{{x}^{2} - 4} \mathrm{dx} = \int \frac{\left({x}^{2} - 4\right) + 4}{\sqrt{{x}^{2} - 4}} \mathrm{dx}$,

$\int \left\{\frac{{x}^{2} - 4}{\sqrt{{x}^{2} - 4}} + \frac{4}{\sqrt{{x}^{2} - 4}}\right\} \mathrm{dx}$,

$= \int \sqrt{{x}^{2} - 4} \mathrm{dx} + 4 \int \frac{1}{\sqrt{{x}^{2} - 4}} \mathrm{dx}$,

$= \left\{\frac{x}{2} \sqrt{{x}^{2} - 4} - \frac{4}{2} \ln | x + \sqrt{{x}^{2} - 4} |\right\} + 4 \ln | x + \sqrt{{x}^{2} - 4} |$.

$I = \frac{x}{2} \sqrt{{x}^{2} - 4} + 2 \ln | x + \sqrt{{x}^{2} - 4} | + {C}_{1}$, as before!

Enjoy Maths.!

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