Chapter 7.2, Problem 21E

To determine

To calculate:

To evaluate the given integral ${\displaystyle \int \frac{dx}{x^2+9}}$ with the initial condition $u=\frac{x}{3}$

Answer to Problem 21E

The solution is $\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+c$

Explanation of Solution

Given information:

The given integral is ${\displaystyle \int \frac{dx}{x^2+9}}$ with the initial condition $u=\frac{x}{3}$

Formula Used:

  ${\displaystyle \int \frac{1}{u^2+1}}du=ta{n}^{-1}\left(u\right)$

Consider the initial condition $u=\frac{x}{3}$ ,

  $u=\frac{x}{3}\Rightarrow 3u=x$

Differentiating with respect to x,

  $du=\frac{1}{3}dx$

  $3du=dx$

Now substituting $3du=dx$ in the integral ${\displaystyle \int \frac{dx}{x^2+9}}$ ,

  ${\displaystyle \int \frac{dx}{x^2+9}={\displaystyle \int \frac{3du}{x^2+9}}}$

Now substituting $3u=x$ in the integral,

  $\begin{array}{l}={\displaystyle \int \frac{3du}{{\left(3u\right)}^2+9}}\\ ={\displaystyle \int \frac{3du}{9u^2+9}}\\ ={\displaystyle \int \frac{3du}{9\left(u^2+1\right)}}\\ =\frac{3}{9}{\displaystyle \int \frac{du}{u^2+1}}\\ =\frac{1}{3}{\displaystyle \int \frac{du}{u^2+1}}\end{array}$

Now using the formula ${\displaystyle \int \frac{1}{u^2+1}}du=ta{n}^{-1}\left(u\right)$ ,

  $=\frac{1}{3}{\tan }^{-1}\left(u\right)+c$

Now substituting back for u,

  $=\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+c$

Therefore, the solution is $\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+c$