List of interesting integrals for early calculus students I am teaching Calc 1 right now and I want to give my students more interesting examples of integrals.  By interesting, I mean ones that are challenging, not as straightforward (though not extremely challenging like Putnam problems or anything).  For example, they have to do a $u$-substitution, but what to pick for $u$ isn't as easy to figure out as it is usually.  Or, several options for $u$ work so maybe they can pick one that works but they learn that there's not just one way to do everything.
So far we have covered trig functions, logarithmic functions, and exponential functions, but not inverse trig functions (though we will get to this soon so those would be fine too).  We have covered $u$-substitution.  Thinks like integration by parts, trig substitution, and partial fractions and all that are covered in Calc 2 where I teach.  So, I really don't care much about those right now.  I welcome integrals over those topics as answers, as they may be useful to others looking at this question, but I am hoping for integrals that are of interest to my students this semester.
 A: I think, in the same nature as Peter Tamaroff's answer, that it is always interesting to learn about the Gamma function:
$$\Gamma(z) := \int_0^\infty e^{-t}t^{z-1}\,\text dt$$
Integration by parts yields
$$\Gamma(z+1) = \left[-e^{-t}t^{z}\right]_0^\infty+z\int_0^\infty e^{-t}t^{z-1}\,\text dt = z\Gamma(z)$$
and seeing that 
$$\Gamma(1) = \int_0^\infty e^{-t}\, dt = \left[-e^{-t}\right]_0^\infty = 1$$
thus for integers $n$:
$$\Gamma(n+1)=n!$$
and we see that the Gamma function is in fact an extension to the factorials for all $z \in \mathbb C$.
A: I think it is great to find a closed form formula for $$\Gamma(n,k)=\int_0^1 (-\log x)^{k-1}x^{n-1} \, dx$$
using integration by parts, where $n,k\in\Bbb N$.
One can prove that
$$\Gamma(n,k)=\int_0^1 (-\log x)^{k-1}x^{n-1}\,dx=\frac{(k-1)!}{n^k}$$
Informally, then, we have that
$$\begin{align}
  \sum\limits_{n = 1}^\infty  \Gamma (n,k) &= \sum\limits_{n = 1}^\infty  \int_0^1 ( - \log x)^{k - 1} x^{n - 1}\,dx  \\
   &= \int_0^1 \frac{( -\log x)^{k - 1}}{1 - x}\,dx \\
   &= (k - 1)!\sum\limits_{n = 1}^\infty \frac{1}{n^k} \\
   &= \Gamma(k)\zeta(k) \end{align} $$
And maybe you can also introduce the Gamma function for integers as $$\Gamma(k) = \int_0^1 ( - \log x)^{k - 1} \, dx $$
which can also be computed by integrations by parts and shown to be $=(k-1)!$. 
Another nice which needs a tad more work, but which is pretty satisfying once is done is $$\int_0^{\pi /2}\sin^m\theta\cos^n\theta \, d \theta=\begin{cases} \dfrac{(m-1)!!(n-1)!!}{(m+n)!!} \text{ if any exponent is odd}\\[10pt] \dfrac{(m-1)!!(n-1)!!}{(m+n)!!} \dfrac{\pi} 2 \text{ both even exponents}  \end{cases}$$
which can be used to prove Wallis's product. 
You can see the details here.
A: I remember having fun with integrating some step functions, for example:
$$\int_{0}^{2} \lfloor x \rfloor - 2 \left\lfloor \frac{x}{2} \right\rfloor \,\mathrm{d}x.$$
My professor for calculus III liked to make us compute piecewise functions, so it would force us to use the Riemann sum definition of the integral.
A: Check out this, this and this and an extended discussion of some useful tricks here
A: Ther area of the circle is a "must have".
$$2\int_{-r}^r\sqrt{r^2-x^2}\,dx.$$
It can be solved in two ways:


*

*by the subsitution $x=r\sin t$ (rather than $r\cos t$), which must be reworked with a trigonometic identity,

*by parts, apparently leading to a dead-end, as after transformation you end-up with the initial integral; but there is an escape...
A: How about the integral of the Standard Normal Distribution? i.e. 
$$ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx = 1  $$
A: You might consider the old warhorse
$$
\int \sec x\ dx
$$
It's very common in calculus texts to resort to the trick of multiplying and dividing by $(\sec x + \tan x)$, upon doing which the answer jumps right out with a bit of simplification. Any reasonable student, though, might complain about this "rabbit out of the hat approach," asking, "How on earth could you expect me to come up with this idea?" All this approach does is impress the student with the author's cleverness while at the same time making them feel stupid. Here's an alternative approach that involves a different, and perhaps more accessible, kind of cleverness.
$$
\begin{align}
\int\sec x\ dx&=\int\frac{1}{\cos x}\ dx=\int \frac{\cos x}{\cos^2 x}\ dx = \int \frac{\cos x}{1-\sin^2 x}\ dx\\
&= \int \cos x\left(\frac{1}{(1-\sin x)(1+\sin x)}\right)\ dx\\
\end{align}
$$
Continue with partial fractions:
$$
\begin{align}
&=\int \frac{\cos x}{2}\left(\frac{1}{1-\sin x}+\frac{1}{1+\sin x}\right)\ dx\\
&= \frac{1}{2}\int\frac{\cos x}{1-\sin x}\ dx+\frac{1}{2}\int \frac{\cos x}{1+\sin x}\ dx\\
\end{align}
$$
and now two simple substitutions and a bit of algebra gives the result. Occasionally, after giving this version I'll give the textbook version as an exercise, where it properly belongs.  
A: Here is an easy integral with a trick.
$$\int \frac{y^2}{(1+y)(1+y^6)} dy$$
The key is to make the substitution $y = 1/x$ and notice that when you're done, after renaming variables, you end up with the (highly similar) integral $$\int \frac{y^3}{(1+y)(1+y^6)} dy$$
Now add the integrals together, and cancel the $1+y$ term, leaving a tractable integral under the further substitution $z = y^3$. Remember that since you've added the integrals, you have to halve whatever answer you get when you're done.
Of course, this could be done with partial fractions also, but that solution is far from elegant. Feel free to make it a definite integral or even an improper integral for additional amusement.
A: A great exercise is to prove that $$\int f^{-1}(x)\mathrm{d}x=xf^{-1}(x)-F\circ f^{-1}(x)+C$$
Where $$F(x)=\int f(x)\mathrm{d}x$$
and $f^{-1}\circ f(x)=f\circ f^{-1}(x)=x$.
This can be done with the substitution $x=f(u)$, and then an integration by parts.
A: One pair of integrals they might find interesting is 
$$\int_0^{\pi/2} \cos^2 x \, dx \textrm{ and } \int_0^{\pi/2} \sin^2 x \, dx.$$
These integrals can be evaluated two different ways.  


*

*Use double angle formulas to find the antiderivatives. 

*Intuitively, the integrals should be the same, because they're the same function only flipped around.  More formally, your students can check that if you make the substitution $u=\frac{\pi}{2}-x$ it turns one integral into the other.  But their sum is  $\int_0^{\pi/2} \sin^2 x + \cos^2 x \, dx=\int_0^{\pi/2} 1 \, dx$. 
By the same trick, you can have your students integrate 
$$\int_0^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx$$
A: I remember spending a lot of time trying to crack $$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}\,dx$$ back in the day, which became much simpler when I found out one could just let $u^6=x$. For your Calc 2 class, I've always been fond of $\int \sqrt{\tan{x}}\,dx$. It uses almost the whole cornucopia of tricks (substitution, completing the square, partial fractions).
Also, sometimes integrals which one might normally approach with trig substitution are much quicker if one knows about explicit formulae for inverse hyperbolic trig functions. For example, $$ \int\frac{1}{\sqrt{x^2+1}}\,dx$$ can be done with a trig substitution, or by noticing this is $\mathrm{arsinh}(x)+c$. In any case, you get $\log{(x+\sqrt{1+x^2}})+ c$, but it just depends whether you'd rather memorize the inverse trig formulae or do the trig subs.
A: Some of my favorite tricks: 
Adding zero
\begin{eqnarray*}
\int \frac{dx}{1+e^x} &=& \int \frac{du}{u(u+1)} 
   \hspace{10ex} (u = e^x) \\
&=& \int du \frac{1+u-u}{u(1+u)} \\
&=& \int \frac{du}{u} - \int \frac{du}{1+u} \\
&=& \ldots
\end{eqnarray*}
Multiplying by one
\begin{eqnarray*}
\int \frac{dx}{1+e^x} &=& \int dx \frac{e^{-x}}{1+e^{-x}} \\
&=& -\int \frac{du}{1+u}
   \hspace{10ex} (u = e^{-x}) \\
&=& \ldots 
%&=& -\log(1+e^{-x}) + C
\end{eqnarray*}
A: You might like
$$ \int \dfrac{dx}{\sqrt{e^{2x}+1}}$$
which, a bit less than obviously, yields to the substitution $u = \sqrt{e^{2x}+1}$.
A: Rick Decker's answer mentioned the "old warhorse": the integral of the secant function.  So I'll take that one a bit further:


*

*There's a Wikipedia article about it: http://en.wikipedia.org/wiki/Integral_of_secant

*It was tabulated by numerical methods in 1599.

*It was first done in 1599 for the purposes of cartography: making maps of large portions of the world.

*The closed form was conjectured in the 1640s.  The conjecture became well known.  Isaac Newton mentioned it.

*It was the first integral ever done by partial fractions (see Rick Decker's answer).

*The problem was solved in the 1660s.



Here's another entertaining integral:
$$
\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx = \frac{22}{7} - \pi.
$$
This integral also has its own Wikipedia article: http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80

You might also consider whether some of the students will be entertained by the tangent half-angle substitution: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution

This one uses integration by parts, so maybe you don't want to use it at this point: http://en.wikipedia.org/wiki/Integral_of_secant_cubed
This one is used in finding the arc length of the parabola and the Archimedean spiral, and the surface area of the helicoid.

(I'm the initial author of all of these articles.  For the first, I relied heavily on an expository paper of V. Frederick Rickey and Philip M. Tuchinsky" "An Application of Geography to Mathematics: History of the Integral of the Secant", Mathematics Magazine, volume 53, number 3, May 1980, pages 162–166.)
A: The following example is interesting because there are several options for a substitution.  This was on a test I was grading and to me one was obvious and it never occurred to me to do any other substitution.  But, the students combined used several choices and to my surprise many worked.
$$\int \sec^8 x \tan x \,dx$$
To me, the obvious choice is $u = \sec x$, $du = \sec x \tan x \,dx$ which leads to
$$\int u^7 \,du = \frac{\sec^8 x}{8} + C$$
But, to my surprise, you can pick other powers of $\sec x$.  If $u = \sec^n x$, where $n$ is a positive integer, then $du = n \sec^n x \tan x \,dx$, so $n = 1, 2, 4, 8$ all work.  For example, if $u = \sec^4 x$, then $du = 4 \sec^4 x \tan x$ and thus we have
$$\frac{1}{4} \int u \,du = \frac{1}{4} \frac{(\sec^4 x)^2}{2} + C$$
A: It's nothing fancy, but I think they are a must show to student to make them see past what they learn.
First one would be after you show them the change of variable technique. At first, student only want to seek what $u$ they can set to cancel something. Have them integrate something of the form
$$
\int x\sqrt{x+1} \,dx.
$$
Letting $u=x+1$ obviously leads to 
$$
\int (u-1)\sqrt{u} \,du
$$
but most student won't know what to do with the $x$ infront of the initial square root.
Another one is right after you show them integration by part. Have them integrate $\ln(x).$
A: I remember one, rather good:
$$\int \dfrac{\text{d}x}{x^4+1}$$
Contrarily to a very similar one where there is a minus sign, this is rather tough but challenging. It starts with partial fractions and substitutions. Give yourself a try, the result is rather elegant:
$$\frac{\log \left(x^2+\sqrt{2} x+1\right)}{4 \sqrt{2}}-\frac{\log \left(x^2-\sqrt{2} x+1\right)}{4 \sqrt{2}}+\frac{\tan ^{-1}\left(\sqrt{2} x+1\right)}{2 \sqrt{2}}-\frac{\tan ^{-1}\left(1-\sqrt{2} x\right)}{2 \sqrt{2}} + c$$
A: I would also recommend the King Property, which is
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
It's a good exercise to prove it as well as to solve some integrals such as
$$\int_{-8}^{10}\frac{\sqrt{\ln(x+9)}}{\sqrt{\ln(x+9)}+\sqrt{\ln(11-x)}}dx$$
$$\int_{0}^{\pi}\frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}+\sqrt{1+\cos x}}dx$$
And the infamous integral
$$\int_{0}^{\pi/2}\frac{1}{1+\tan^{2022}x}dx$$
