Daily Calculus - Post one Problem and its Solution

[Eli]

Calculus is a natural witchcraft to many problems in life, it is highly applied to solve a wide range of problems in sciences, engineering, mathematics, humanities, economics, and many other areas. In this Daily Calculus thread, I encourage anyone to post at least a problem and its solution to help others learn faster.

Problem 1.

Evaluate
\begin{align}
\int \frac{dx}{\sqrt{1 - x^2}}.
\end{align}

Solution:

Let's apply trigonometric substitution
\begin{align}
x = {\rm sin}u, \implies {\rm d} x = {\rm cos}u{\rm d}u.
\end{align}
Then
\begin{align}
\begin{split}
\int \frac{dx}{\sqrt{1 - x^2}} &\quad = \int \frac{{\rm cos}u{\rm d}u}{\sqrt{1 - {\rm sin}^{2}u}}\\
\\
&\quad = \int \frac{{\rm cos}u{\rm d}u}{\sqrt{{\rm cos}^{2}u}} \ ({\rm since } \ \ {\rm cos}^{2}u + {\rm sin}^{2}u = 1)\\
\\
&\quad = \int \frac{{\rm cos}u{\rm d}u}{{\rm cos}u}\\
\\
&\quad = \int{\rm d}u = u + c = {\rm sin}^{-1}x + c
\end{split}
\end{align}
where \(c\) is a constant.

Problem 2.

Evaluate
\begin{align}
\int\frac{{\rm d}x}{\sqrt{1 - ax^{2}}}.
\end{align}

Solution:

By the analogy of the solution of the integral above,
\begin{align}
\int\frac{{\rm d}x}{\sqrt{1 - ax^{2}}} = \frac{{\sin}^{-1}(\sqrt{a}x)}{\sqrt{a}} + c. \ \ ({\rm Hint}: \ {\rm Make \ substitution} \ \sqrt{a}x = {\rm sin}u.)
\end{align}

[Forbidden_Technology]

Are you ready to solve this calculus problem?

[Eli]

Sage or Maxima (see here) yields

\begin{equation}
\int\frac{1}{1-x^8}{\rm d}x = 1/8\sqrt{2}\times {\rm tan}^{-1}\left(1/2 \sqrt{2}\times (2x + \sqrt{2})\right) + 1/8 \sqrt{2}\times {\rm tan}^{-1}\left(1/2 \sqrt{2}\times(2x - \sqrt{2})\right) + 1/16 \sqrt{2}\times{\rm log}(x^2 + \sqrt{2}\times x + 1) - 1/16\sqrt{2}\times{\rm log}(x^2 - \sqrt{2}\times x + 1) + 1/4 {\rm tan}^{-1}(x) + 1/8 {\rm log}(x + 1) - 1/8 {\rm log}(x - 1)
\end{equation}