[SOLVED] Trig Substitution Integral

**Paperwings** · December 20th 2009, 08:24 AM

Evaluate $\int \frac{1}{x^2\sqrt{x^2+4}}$
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Using trig substitution, let $x=2\tan{u}$ , $dx=2\sec^2{u}du$

$= \int \frac{1}{4\tan^2{u} \cdot \sqrt{4\tan^2{u}+4}} \cdot 2\sec^2{u}du$

$= \int \frac{2\sec^2{u}}{\left( 4\tan^2{u} \right) \sqrt{4 ( \tan^2{u}+1) }}du$

$= \int \frac{2\sec^2{u}}{\left( 4\tan^2{u} \right) 2 \sqrt{\tan^2{u}+1}}du$

Since $\sqrt{\tan^2{u}+1}= \sec{u}$ , then

$= \int \frac{\sec^2{u}}{4\tan^2{u}\sec{u}} du$

$= \frac{1}{4} \int \frac{\sec{u}}{\tan^2{u}} du$

$= \frac{1}{4} \int \frac{\csc{u}}{\tan{u}} du$

From here, I do not know how to proceed.

$\csc{u} = \frac{1}{\sin{u}}$
$\tan{u} = \frac{\sin{u}}{\sin{u}}$

$= \frac{1}{4} \int \frac{\cos{u}}{\sin^2{u}} du$

$= \frac{1}{4} \int \frac{\cos{u}}{1-\cos^2{u}} du$

**red_dog** · December 20th 2009, 08:30 AM

$\frac{\csc u}{\tan u}=\frac{\cos u}{\sin^2u}$

Let $\sin u=t$

**Soroban** · December 20th 2009, 10:35 AM

Hello, Paperwings!

You have: . $\frac{1}{4} \int \frac{\csc{u}}{\tan{u}}\,du \;=\;\frac{1}{4}\int \csc u \cot u\,du \;=\;-\frac{1}{4}\csc u + C$

**Paperwings** · December 20th 2009, 08:56 PM

Ah, yes. Thank you. So,

$= \frac{1}{4} \int \frac{\cos{u}}{\sin^2{u}} du$

Let $\sin{u} = t$ , $\cos{u} = dt$

$= \frac{1}{4} \int \frac{1}{t^2}dt$

$= - \frac{1}{4t} + C$

$= - \frac{1}{4\sin{u}} + C = - \frac{1}{4}\csc{u} + C$

Since $x = 2\tan{u}$

$= - \frac{1}{4} \csc{\arctan{\frac{x}{2}}} + C$

I am a little unsure how to simplify this.

I picture a triangle with sides x and 2. the hypothenuse is $\sqrt{x^2+4}$ .

Ah, ok. it's

$= - \frac{1}{4} \cdot \frac{\sqrt{x^2+4}}{x} + C$

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