Table Integral: \int \frac{dx}{x^2 \sqrt{a^2 - x^2}} = - \frac{\sqrt{a^2 - x^2}}{a^2 x} + c Use...

Question:

Table Integral:

{eq}\int \frac{dx}{x^2 \sqrt{a^2 - x^2}} = - \frac{\sqrt{a^2 - x^2}}{a^2 x} + c {/eq}

Use Table Integral to evaluate the integral. (Do not forget about substitution):

{eq}\int \frac{dx}{x^2 \sqrt{16 - 9x^2}} ={/eq}

Integration:

Given :

{eq}\int \frac{dx}{x^2 \sqrt{a^2 - x^2}} = - \frac{\sqrt{a^2 - x^2}}{a^2 x} + c {/eq}

To Find : {eq}\int \frac{dx}{x^2 \sqrt{16 - 9x^2}} {/eq}

We can write {eq}16 - 9x^2=9\left ( \frac{16}{9}-x^2 \right ) {/eq}

We will use above given formula to find the required answer .

Answer and Explanation:

{eq}\begin{align*} \displaystyle \int \frac{dx}{x^2 \sqrt{16 - 9x^2}} &=\int \frac{dx}{x^2 \sqrt{9\left ( \frac{16}{9} - x^2 \right )}}\\ \displaystyle &=\frac{1}{3}\int \frac{dx}{x^2 \sqrt{\left ( \frac{16}{9} - x^2 \right )}}\\ \displaystyle &=\frac{1}{3}\int \frac{dx}{x^2 \sqrt{\left ( \left ( \frac{4}{3} \right )^2 - x^2 \right )}}\\ \end{align*} {/eq}

We will use formula : {eq}\int \frac{dx}{x^2 \sqrt{a^2 - x^2}} = - \frac{\sqrt{a^2 - x^2}}{a^2 x} + c {/eq}

Take {eq}a=\frac{4}{3} {/eq}

So ,

{eq}\begin{align*} \displaystyle \int \frac{dx}{x^2 \sqrt{16 - 9x^2}} &=\frac{1}{3}\int \frac{dx}{x^2 \sqrt{\left ( \left ( \frac{4}{3} \right )^2 - x^2 \right )}}\\ \displaystyle &= \frac{-1}{3} \frac{\sqrt{\left ( \frac{4}{3} \right )^2 - x^2}}{\left ( \frac{4}{3} \right )^2 x} + c\\ \displaystyle &=\frac{-1}{3}\left ( \frac{9}{16} \right )\frac{1}{x}\sqrt{\frac{16}{9}-x^2}+C\\ \displaystyle &=\frac{-1}{16x}\sqrt{16-9x^2}+C\\ \end{align*} {/eq}


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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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