Single Variable Calculus, Chapter 8, 8.3, Section 8.3, Problem 36

Evaluate $\displaystyle \int \frac{dx}{x^4 \sqrt{x^2-2 }}$, check your answer if its reasonable by grouping the integrand and its indefinite integral on the same screen.
By trigonometric substitution,



$\displaystyle\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{2}}{x}$ and $\displaystyle \tan \theta = \frac{\sqrt{x^2 - 2}}{\sqrt{2}}$
$\displaystyle x = \frac{\sqrt{2}}{\cos \theta} = \sqrt{2} \sec \theta$

So,

$
\begin{equation}
\begin{aligned}
\int \frac{dx}{x^4 \left( \sqrt{x^2 - 2} \right) } &= \int \frac{\sqrt{2}\sec \theta \tan \theta d \theta}{ \left( \sqrt{2} \sec \theta \right)^4 ( \sqrt{2} \tan \theta )}\\
\\
&= \int \frac{d \theta}{4 \sec^3 \theta}\\
\\
&= \frac{1}{4} \int \cos^3 \theta d \theta
\end{aligned}
\end{equation}
$

By using integration by parts,
If we let $u = \cos^2 \theta$ and $dv = \cos \theta d \theta$, then
$du = 2 \cos \theta ( - \sin \theta d \theta)$ and $\displaystyle v = \int \cos \theta d\theta = \sin \theta$

So,
$\displaystyle \int \cos^3 \theta d \theta = uv - \int v du = \sin \theta \cos^2 \theta + 2 \int \sin^2 \theta \cos \theta d \theta$


$
\begin{equation}
\begin{aligned}
\text{by substituting, if we let } z &= \sin \theta \text{ , then } dz = \cos \theta d \theta\\
\\
&= \sin \theta \cos^2 \theta + 2 \int z^2 dz\\
\\
&= \sin \theta \cos^2 \theta + 2 \left[ \frac{z^3}{3} \right]\\
\\
&= \sin \theta \cos^2 \theta + 2 \frac{(\sin \theta)^3}{3}
\end{aligned}
\end{equation}
$


So,
$\displaystyle \frac{1}{4} \int \cos^3 \theta d \theta = \frac{1}{4} \left[ \sin \theta \cos^2 \theta + \frac{2(\sin \theta)^3}{3}\right] + c$

From the triangle,

$
\begin{equation}
\begin{aligned}
&= \frac{1}{4} \left[ \left( \frac{\sqrt{x^2-2}}{x} \right) \left( \frac{\sqrt{2}}{2} \right)^2 + \frac{2}{3} \left( \frac{\sqrt{x^2-2}}{x} \right)^2 \right]\\
\\
&= \frac{1}{4} \left[ \frac{2\sqrt{x^2-2}}{x^3} + \frac{2}{3} \left( \frac{\sqrt{x^2 - 2}}{x^3} \right)^3 \right]\\
\\
&= \left[ \frac{\sqrt{x^2 - 2}}{2x^3} + \frac{\left( \sqrt{x^2 - 2 } \right)^3}{6x^3} \right]\\
\\
\int \frac{dx}{x^4 \sqrt{x^2-2}} &= \frac{3\sqrt{x^2 - 2}+\left( \sqrt{x^2 - 2} \right)^3}{6x^3} + c\\
\end{aligned}
\end{equation}
$




Based from the graph, we can say that our answer is reasonable because $f$ is increasing whenever $f$ is positive. On the other hand, $f$ is decreasing whenever $f'$ is negative.