# Equivalence of two antiderivatives involving trigonometric/hyperbolic functions

I am struggling to see how two antiderivatives of the same function—obtained in two different ways—are equivalent (what I mean by equivalent is that they differ from just a constant), if they even are equivalent.

The function in question is

\begin{align} f \colon \quad \{x\in\mathbb{R} \mid e^{2x} \ge 9\} &\to\mathbb{R}\\ x &\mapsto f(x) = \frac{1}{\sqrt{e^{2x} - 9}} \end{align}\,.

Wanting to find $$I = \displaystyle\int f(x)\mathrm{d}x$$, I proceeded as follows.

$$\int \frac{1}{\sqrt{e^{2x} - 9}}\mathrm{d}x = \frac{1}{3} \int \frac{1}{\sqrt{\left(\dfrac{e^x}{3}\right)^2 - 1}}\mathrm{d}x\,.$$

Since $$\left(\frac{e^x}{3}\right)^2 \in[1,+\infty]$$ and the image of $$\cosh$$ is $$[1, +\infty)$$, $$\exists t \in \mathbb{R}\colon\quad \left(\dfrac{e^x}{3}\right)^2 = \cosh^2t\\ \Rightarrow \mathrm{d}x = \tanh t\ \mathrm{d}t\,.$$

Thus,

$$I = \frac{1}{3} \int \frac{\tanh t}{\sqrt{\cosh^2 t - 1}}\mathrm{d}t = \frac{1}{3}\int \mathrm{sech}\ t\ \mathrm{d}t\,.$$

Here is where the two different ways diverge. The first thing I tried was using the hyperbolic function's definition.

$$I = \frac{1}{3}\int \frac{2e^t}{e^{2t} + 1}\ \mathrm{d}t\,; \qquad\text{let}\quad \begin{cases} u = e^t\\ \mathrm{d}u = e^t\mathrm{d}t \end{cases}\\ \\ \Rightarrow I = \frac{2}{3} \int \frac{1}{u^2 + 1}\ \mathrm{d}u = \frac{2}{3} \arctan u + C = \frac{2}{3} \arctan e^t + C = \\ = \frac{2}{3} \arctan e^{\text{arccosh} \frac{e^x}{3}} + C = \frac{2}{3} \arctan e^{\ln\left(\frac{e^x + \sqrt{e^{2x}-9}}{3}\right)} + C =\\ = \frac{2}{3} \arctan \left(\frac{e^{x} + \sqrt{e^{2x}-9}}{3}\right) + C\,.$$

The alternative solution is

$$I = \frac{1}{3}\int \mathrm{sech}\ t\ \mathrm{d}t = \frac{1}{3}\int \frac{\cosh t}{\cosh^2 t}\ \mathrm{d}t = \frac{1}{3}\int \frac{\cosh t}{1 +\sinh^2 t}\ \mathrm{d}t =\\ = \frac{1}{3}\arctan \left(\sinh t\right) + C = \frac{1}{3}\arctan \left(\sinh \left(\mathrm{arccosh} \frac{e^x}{3}\right)\right) + C = \frac{1}{3}\arctan \frac{\sqrt{e^{2x} - 9}}{3} + C\,.$$

I don't see any obvious way in which $$\frac{2}{3} \arctan \left(\frac{e^x + \sqrt{e^{2x}-9}}{3}\right) + C$$ and $$\frac{1}{3}\arctan \frac{\sqrt{e^{2x} - 9}}{3} + C$$ are equivalent antiderivatives, although they do seem to be. # Update

Following Yuriy S's suggestion, I tried the following. Let $$s = \frac{\sqrt{e^{2x} - 9}}{3}\,.$$ Thus, $$\arctan s = 2\arctan \frac{\sqrt{1+s^2}-1}{s} = 2\arctan \frac{\sqrt{1+\frac{e^{2x}-9}{9}}-1}{\frac{\sqrt{e^x - 9}}{3}} = 2\arctan \frac{3e^x-3}{\sqrt{e^x - 9}}\,.$$

On the other hand, let $$r = \frac{e^x + \sqrt{e^{2x} - 9}}{3}\,,$$ so that $$\frac{1}{r} = \frac{3}{e^x + \sqrt{e^{2x} - 9}} \cdot \frac{e^x - \sqrt{e^{2x} - 9}}{e^x - \sqrt{e^{2x} - 9}} = \frac{e^x - \sqrt{e^{2x} - 9}}{3}\,.$$

I was striving to apply the formula $$\arctan x + \arctan \frac{1}{x} = \pm\frac{\pi}{2}$$ but I got stuck here.

• Maybe this formula might help: $$\arctan x=2 \arctan \frac{\sqrt{1+x^2}-1}{x}$$ – Yuriy S Dec 29 '18 at 16:22

Whenever you face this problem, you can check the answer by differentiating. If $$F(x)=\frac{2}{3} \arctan \left(\frac{e^{x} + \sqrt{e^{2x}-9}}{3}\right) + C$$ then \begin{align} F'(x) &=\frac{2}{3}\frac{1}{1+\left(\dfrac{e^{x} + \sqrt{e^{2x}-9}}{3}\right)^2} \frac{1}{3}\left(e^x+\frac{e^{2x}}{\sqrt{e^{2x}-9}}\right) \\[4px] &=\frac{2}{9+e^{2x}+e^{2x}-9+2e^x\sqrt{e^{2x}-9}} \frac{e^x(e^x+\sqrt{e^{2x}-9}\,)}{\sqrt{e^{2x}-9}} \\[4px] &=\frac{1}{\sqrt{e^{2x}-9}} \end{align} We can try differentiating the second function $$G(x)=\frac{1}{3}\arctan\frac{\sqrt{e^{2x}-9}}{3}+C$$ and we get \begin{align} G'(x) &=\frac{1}{3}\frac{1}{1+\left(\dfrac{\sqrt{e^{2x}-9}}{3}\right)^2} \frac{1}{3}\frac{e^{2x}}{\sqrt{e^{2x}-9}} \\[4px] &=\frac{1}{9+e^{2x}-9}\frac{e^{2x}}{\sqrt{e^{2x}-9}} \\[4px] &=\frac{1}{\sqrt{e^{2x}-9}} \end{align}
You can easily compute the constant difference by the limits at $$\infty$$ (with $$C=0$$ in both cases): $$\lim_{x\to\infty}F(x)=\frac{2}{3}\frac{\pi}{2}=\frac{\pi}{3} \qquad \lim_{x\to\infty}G(x)=\frac{1}{3}\frac{\pi}{2}=\frac{\pi}{6}$$
Alternative solution: substitute $$\sqrt{e^{2x}-9}=3t$$, so $$x=\frac{1}{2}\log(9(t^2+1))$$ and $$dx=\frac{t}{t^2+1}\,dt$$ so the integral becomes $$\int\frac{1}{t^2+1}\,dt=\arctan t+C=\arctan\frac{\sqrt{e^{2x}-9}}{3}+C$$
• @Anakhand Compute the limits at $\infty$. – egreg Dec 29 '18 at 17:48