# Evaluate the integral: {eq}\displaystyle \int_1^2 \frac {1} {x^2 \sqrt {4 x^2 + 9}}\ dx {/eq}.

## Question:

Evaluate the integral:

{eq}\displaystyle \int_1^2 \frac {1} {x^2 \sqrt {4 x^2 + 9}}\ dx {/eq}.

## Integration by Trigonometric Substitution:

Definition: For the radical expressions which cannot be solved directly, we use the trigonometric substitution which converts the integral into a form that can be easily solve by simple integration methods.

For example, an integration of the form

{eq}\displaystyle \int \frac{1}{\sqrt{b^2- u^2}}. {/eq}

which we cannot solve using direct integration methods.

So, apply trigonometric substitution {eq}\displaystyle u = b \sin v. {/eq}

Now, we can solve it directly by simple integration mtheod.

Some related concepts that are useful in conjunction with this concept:

1. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\tan \left(x\right)\right)=\sec ^2\left(x\right). {/eq}

2. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\sin \left(x\right)\right)=\cos \left(x\right). {/eq}

3. Take the constant out from the integration: {eq}\displaystyle \int a\cdot g\left(y\right)dy=a\cdot \int g\left(y\right)dy {/eq}

4. The power rule: {eq}\displaystyle \int x^adx=\frac{x^{a+1}}{a+1}, \quad a\ne -1. {/eq}

5. The exponent property: {eq}\displaystyle \frac{1}{v^2}=v^{-2}. {/eq}

## Answer and Explanation: 1

We have to find the integration of $$\displaystyle I = \int_1^2 \frac {1} {x^2 \sqrt {4 x^2 + 9}}\ dx$$

Apply the trigonometric substitution {eq}\displaystyle x=\frac{3}{2}\tan \left(u\right) \Rightarrow dx = \frac{3}{2} \sec ^2\left(u\right) du. {/eq}

Limits: {eq}1 \rightarrow \arctan \left(\frac{2}{3}\right) {/eq} and {eq}2 \rightarrow \arctan \left(\frac{4}{3}\right) {/eq}

$$\displaystyle = \int _{\arctan \left(\frac{2}{3}\right)}^{\arctan \left(\frac{4}{3}\right)}\frac{2\sec \left(u\right)}{9\tan ^2\left(u\right)}du$$

Take the constant out from the integration.

$$\displaystyle = \frac{2}{9}\cdot \int _{\arctan \left(\frac{2}{3}\right)}^{\arctan \left(\frac{4}{3}\right)}\frac{\sec \left(u\right)}{\tan ^2\left(u\right)}du$$

Express with sin and cos.

\begin{align*} \displaystyle &= \frac{2}{9}\cdot \int _{\arctan \left(\frac{2}{3}\right)}^{\arctan \left(\frac{4}{3}\right)}\frac{\frac{1}{\cos \left(u\right)}}{\left(\frac{\sin \left(u\right)}{\cos \left(u\right)}\right)^2}du\\ \displaystyle &= \frac{2}{9}\cdot \int _{\arctan \left(\frac{2}{3}\right)}^{\arctan \left(\frac{4}{3}\right)}\frac{\cos \left(u\right)}{\sin ^2\left(u\right)}du \end{align*}

Again apply the trigonometric substitution for {eq}\displaystyle v=\sin \left(u\right) \Rightarrow dv = \cos \left(u\right) du. {/eq} Also using {eq}sin(arctanx)=|x|/\sqrt{(1+x^2)} {/eq}

Limits: {eq}\arctan \left(\frac{2}{3}\right) \rightarrow \frac{2\sqrt{13}}{13} {/eq} and {eq}\arctan \left(\frac{4}{3}\right) \rightarrow \frac{4}{5}. {/eq}

$$\displaystyle = \frac{2}{9}\cdot \int _{\frac{2\sqrt{13}}{13}}^{\frac{4}{5}}\frac{1}{v^2}dv$$

Use the exponent property.

$$\displaystyle = \frac{2}{9}\cdot \int _{\frac{2\sqrt{13}}{13}}^{\frac{4}{5}}v^{-2}dv$$

Apply the power rule.

$$\displaystyle = \frac{2}{9}\left[-\frac{1}{v}\right]^{\frac{4}{5}}_{\frac{2}{\sqrt{13}}}$$

Compute the boundaries.

\begin{align*} \displaystyle &= \frac{2}{9}\left[-\frac{5}{4}-\left(-\frac{\sqrt{13}}{2}\right)\right]\\ \displaystyle &= \frac{2}{9}\left(-\frac{5}{4}+\frac{\sqrt{13}}{2}\right)\\ \displaystyle &= \frac{-5+2\sqrt{13}}{18}. \end{align*}