# Evaluate the integral using the indicated trigonometric substitute. (Use C for the constant of...

## Question:

Evaluate the integral using the indicated trigonometric substitute. (Use {eq}C {/eq} for the constant of integration.)

{eq}\displaystyle\int \frac{x^3}{\sqrt{x^2 + 36}} \, \mathrm{d}x, \: x = 6 \tan \theta {/eq}

Sketch and label the associated right triangle.

## Trigonometric Substitution:

When integrating a function that contains terms of the form {eq}\sqrt{x^2 + a^2}, {/eq} we can make the trigonometric substitution {eq}x = a \tan \theta. {/eq} Then {eq}dx = a \sec^2 \theta \: d\theta, {/eq} and

{eq}\sqrt{x^2 + a^2} = \sqrt{(a \tan \theta)^2 + a^2} = \sqrt{a^2 (\tan^2 \theta + 1)} = \sqrt{a^2 \sec^2 \theta} = a \sec \theta. {/eq}

This eliminates the square root.

We are given {eq}x = 6 \tan \theta. {/eq} Taking the derivative gives {eq}dx = 6 \sec^2 \theta \: d\theta. {/eq} Now we can substitute these into the integral:

{eq}\begin{eqnarray*}\int \displaystyle\frac{x^3}{\sqrt{x^2 + 36}} \, \mathrm{d}x & = & \int \displaystyle\frac{(6 \tan \theta)^3}{\sqrt{(6 \tan \theta)^2 + 36}} \: (6 \sec^2 \theta) \: d\theta \\ \\ & = & \int \displaystyle\frac{216 \tan^3 \theta}{\sqrt{36 \tan^2 \theta + 36}} \: 6 \sec^2 \theta \: d\theta \\ \\ & = & \int \displaystyle\frac{216 \tan^3 \theta}{\sqrt{36 (\tan^2 \theta + 1)}} \: 6 \sec^2 \theta \: d\theta \\ \\ & = & \int \displaystyle\frac{216 \tan^3 \theta}{\sqrt{36 \sec^2 \theta}} \: 6 \sec^2 \theta \: d\theta \\ \\ & =& \int \displaystyle\frac{216 \tan^3 \theta}{6 \sec \theta} \: 6 \sec^2 \theta \: d\theta \\ \\ & = & 216 \int \tan^3 \theta \sec \theta \: d\theta \\ \\ & = & 216 \int \tan^2 \theta \tan \theta \sec \theta \: d\theta \\ \\ & = & 216 \int (\sec^2 \theta - 1) \sec \theta \tan \theta \: d\theta \end{eqnarray*} {/eq}

Now, let {eq}u = \sec \theta. {/eq} Then {eq}du = \sec \theta \tan \theta \: d\theta. {/eq} This gives

{eq}\begin{eqnarray*}216 \int (\sec^2 \theta - 1) \sec \theta \tan \theta \: d\theta & = & 216 \int (u^2 - 1) \: du \\ & =& 216 \left( \displaystyle\frac{u^3}3 - u \right) + C \\ & = & 72 \sec^3 \theta - 216 \sec \theta + C \end{eqnarray*} {/eq}

Since {eq}x = 6 \tan \theta, {/eq} we have {eq}\tan \theta = \displaystyle\frac{x}{6} = \frac{\mathrm{opposite}}{\mathrm{adjacent}}. {/eq} Consider the following triangle:

Therefore {eq}\sec \theta = \displaystyle\frac{\mathrm{hypotenuse}}{\mathrm{adjacent}} = \frac{\sqrt{x^2 + 36}}{6}, {/eq} and

{eq}\begin{eqnarray*}\displaystyle\int \frac{x^3}{\sqrt{x^2 + 36}} \, \mathrm{d}x & = & 72 \sec^3 \theta - 216 \sec \theta + C \\ & = & 72 \left( \displaystyle\frac{\sqrt{x^2 + 36}}{6} \right)^3 - 216 \left( \displaystyle\frac{\sqrt{x^2 + 36}}{6} \right) + C \\ & = & \displaystyle\frac{(x^2 + 36)^{3/2}}{3} - 72 (x^2 + 36)^{1/2} + C \end{eqnarray*} {/eq}