(35S) Quick integration

[Thomas Klemm]

(11-19-2022 01:42 AM)Liamtoh Resu Wrote:  Can someone walk me through the symbolic calculation of the function?

I used the substitution:

\(
\begin{align}
2x &= \sinh(u) \\
4x^2 &= \sinh^2(u) \\
1 + 4x^2 &= 1 + \sinh^2(u) \\
&= \cosh^2(u) \\
\sqrt{1 + 4x^2} &= \cosh(u) \\
\end{align}
\)

After applying \(\frac{d}{du}\) on both sides of the substitution we get:

\(
\begin{align}
2\frac{dx}{du} &= \cosh(u) \\
\end{align}
\)

From this:

\(
\begin{align}
dx &= \tfrac{1}{2}\cosh(u) \, du \\
\end{align}
\)

Replace the substitution in the original integral:

\(
\begin{align}
\int \sqrt{1 + 4 x^2} \, dx
&= \tfrac{1}{2} \int \cosh^2(u) \, du \\
&= \tfrac{1}{4} \int 1 + \cosh(2u) \, du \\
&= \tfrac{1}{4} \left( u + \tfrac{1}{2} \sinh(2u) \right) +c \\
&= \tfrac{1}{8} \left( 2u + \sinh(2u) \right) +c \\
\end{align}
\)

Here we used the double angle formula:

\(
\begin{align}
\cosh(2u) = 2\cosh^2(u) - 1
\end{align}
\)

Now we can substitute the lower and upper limits:

\(
\begin{align}
x = 0 \Rightarrow 2x = 0 &= \sinh(u) \rightarrow u = 0 \\
x = 1 \Rightarrow 2x = 2 &= \sinh(u) \rightarrow u = \sinh^{-1}(2) \\
\end{align}
\)

Plugging them into the antiderivative leads to:

\(
\begin{align}
\int_0^1 \sqrt{1 + 4 x^2} \, dx
&= \frac{2 \sqrt{5} + \sinh^{-1}(2)}{4} \\
\end{align}
\)

Here again we used another double angle formula:

\(
\begin{align}
\sinh(2u)
&= 2\sinh(u)\cosh(u) \\
&= 2 \cdot 2x \cdot \sqrt{1 + 4x^2}
\end{align}
\)

---

However e.g. WolframAlpha comes up with a different substitution:

\(
\begin{align}
x &= \frac{\tan(u)}{2} \\
\end{align}
\)

But I'm too lazy to write that down here.
Maybe you want to give it a try?

References

• Double-Angle Formulas