What are other unexpected results of integration?

Question

I have integral of $\dfrac{1}{t^2 + 1}$ and integral of $\dfrac{t}{t^2 + 1}$ whose output is $\arctan(t)$ and $\dfrac12\ln(t^2 + 1)$ respectively.

Are there any similar unexpected results when we take integral? A link to such list would be enough for me. I'm asking this because while solving heat equations in differential equations I always face similar things. Thus I need to be prepared when I face something similar where the result of the integral is tangent, sine, cosine, permutation, cot, power function, etc.

Answer 1

I wouldn't say that either of these results is unexpected, but I agree that inverse $\tan$ wouldn't spring to mind to somebody who only knew the power rule for integration.

Let's say we've got $\int \frac{1}{\color{red}{1+t^2}}\color{green}{dt}$.

Let $t=\tan(\theta) \iff \color{purple}{\theta=\arctan(t)}$.

Now, $\underbrace{\frac{dt}{d\theta}=\sec^2(\theta)}_{\textrm{try differentiating tan by the quotient rule!}} \iff dt=\color{green}{\sec^2(\theta)d\theta}$

Then the integral becomes $$\int \frac{1}{\color{red}{1+\tan^2(\theta)}}\color{green}{\sec^2(\theta) d\theta}=\require{cancel} \int\frac{1}{{\cancel{\color{red}{\sec^2(\theta)}}}}\cancel{\color{green}{\sec^2(\theta)}}\color{green}{d\theta}=\int 1d\theta=\color{purple}{\theta}+C=\color{purple}{\arctan(t)}+C.$$

For the second integral, use the substitution $u=t^2+1.$ I'll leave you to prove the second result!

By the way, in line with the whole 'unexpected integrals topic', for me, at least: $$\int \frac{1}{x}dx =\ln|x|+C$$ (unintuitive because $\frac{1}{x}$ doesn't obey the power rule like every other power does and $\frac{1}{x}$ doesn't seem to be related to $\ln$ in any way).

A couple more, following your pattern of unintuitiveness, are:

• $\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin(x)+C$
• $\int \frac{1}{\sqrt{1+x^2}}dx=\mathrm{arsinh}(x)+C$
• $\int \frac{1}{\sqrt{x^2-1}}dx=\mathrm{arcosh}(x)+C$
• $\int \frac{1}{1-x^2}dx=\frac{1}{2}\ln \left| \frac{x+1}{x-1}\right|+C$ (for $x \neq \pm1$).

Answer 2

Let me illustrate the case you give at the beginning of the post. Each function will be developed as an infinite series built at the origin. So, $$\dfrac{1}{t^2 + 1}=1-t^2+t^4-t^6+t^8-t^{10}+t^{12}+O\left(t^{13}\right)$$ from which we can derive (just multiply each term by $t$) $$\dfrac{t}{t^2 + 1}=t-t^3+t^5-t^7+t^9-t^{11}+t^{13}+O\left(t^{14}\right)$$ and the functions "look" similar. But now, let us integrate them and get $$\int \dfrac{dt}{t^2 + 1}=t-\frac{t^3}{3}+\frac{t^5}{5}-\frac{t^7}{7}+\frac{t^9}{9}-\frac{t^{11}}{11}+\frac{t^{ 13}}{13}+O\left(t^{14}\right)$$ $$\int \dfrac{t ~~dt}{t^2 + 1}=\frac{t^2}{2}-\frac{t^4}{4}+\frac{t^6}{6}-\frac{t^8}{8}+\frac{t^{10}}{10}-\frac{t^{12 }}{12}+\frac{t^{14}}{14}+O\left(t^{15}\right)$$ After integration, they really don't look any more at all to each other. So, there is no reason for the integrals to look similar.

However, you could find some similarities in the following integrals $$I_n=\int \dfrac{t^{2n}}{t^2 + 1}~~dt$$ $$J_n=\int \dfrac{t^{2n-1}}{t^2 + 1}~~dt$$ since $$I_0=\tan ^{-1}(t)$$ $$I_1=t-\tan ^{-1}(t)$$ $$I_2=\frac{t^3}{3}-t+\tan ^{-1}(t)$$ $$I_3=\frac{t^5}{5}-\frac{t^3}{3}+t-\tan ^{-1}(t)$$ and $$J_1=\frac{1}{2} \log \left(t^2+1\right)$$ $$J_2=\frac{t^2}{2}-\frac{1}{2} \log \left(t^2+1\right)$$ $$J_3=\frac{1}{4} \left(\left(t^2-1\right)^2+2 \log \left(t^2+1\right)\right)$$

Answer 3

There is a great deal of similarity between the two results:

$\begin{align} &\displaystyle \int \dfrac{t}{t^2 + 1} dt = \dfrac{\ln (t^2 + 1)}{2} + C = \dfrac{\ln(t-i)(t + i)}{2} + C &= \boxed{\dfrac{\ln(t - i) + \ln(t + i)}{2} + C}\\ &\displaystyle \int \dfrac{1}{t^2 + 1} dt = \arctan t + C = \dfrac{1}{2i}\ln\dfrac{t-i}{t+i} + C & = \boxed{\dfrac{\ln(t - i) - \ln(t + i)}{2i} + C} \end{align}$

Tell me those aren't similar!