Today, scrolling through Instagram, I came across a post from the user pringlesmaths that caught my attention.
I am quite a fan of his/her publications, since they are papers featuring interesting problems from different Olympiads; in this case it was an integral
from the MIT Integration Bee 2010 which was surprisingly easy to solve, and as I delved into the other editions of the contest I found many others that I
believe any high-schooler could do after thinking for some time.
Before proceeding, though, I must mention that, first, these integrals all come from the qualifying rounds rather than the actual, much more challenging
competition, and second, the real difficulty comes from solving them in a very short period of time – students are given only 20 minutes to complete
the whole test –. That being said, at least the ones I will discuss in a moment were almost immediate to a relatively untrained eye such as mine.
Let us start with the one in pringlesmaths' sheet; he/she used the Laplace transform in the solution, but it can be done in a much easier fashion:
Evaluate the integral
$$
{\rm I} = \lint_0^1 \sin^2 (\log x) \diff x.
$$
Proposed in MIT Integration Bee Qualifying Exam, 2010, Q5.
Notice that substituting $t := \log x$ transforms the integral into the product of an exponential and a sine squared, which by the double-angle cosine formula
can be further simplified into an exponential times a cosine. These integrals are well-known examples of the so-called cyclic integrals, and their result is
revealed after two iterations of integration by parts by leveraging the fact that sines are cosines acquire a minus sign after being differentiated twice:
$$
\lint_0^1 \sin^2 (\log x) \diff x = \left\{\begin{aligned}
t &= \log x\\
\diff t &= \diff x/x
\end{aligned}\right\} = \lint_{-\infty}^0 e^t \sin^2 t \diff t
= \frac12 \lint_{-\infty}^0 e^t (1 - \cos(2t)) \diff t = \frac12 - \frac12\underbrace{\lint_{-\infty}^0 e^t \cos(2t) \diff t}_{\rm J}.
$$
As said, performing integration by parts on $\rm J$ and the subsequent $\rm K$ yields
$$
{\rm J} = e^t \cos(2t)\bigg|_{-\infty}^0 + 2 \underbrace{\lint_{-\infty}^0 e^t \sin(2t) \diff t}_{\rm K};
\qquad {\rm K} = \cancel{e^t \sin(2t)\bigg|_{-\infty}^0} - 2\lint_{-\infty}^0 e^t \cos (2t) \diff t = -2{\rm J},
$$
from which we can solve for $\rm J$ and then substitute back into $\rm I$ to finish off the problem
$$
{\rm J} = 1 - 4{\rm J} \implies {\rm J} = \frac15 \implies {\rm I} = \frac12 - \frac1{10} = \color{green}\boxed{\frac25.}
$$
We stay in 2010 for the next one, which is even shorter:
Find the following antiderivative
$$
\int \left(\frac1{\log x} - \frac1{\log^2 x}\right) \diff x.
$$
Proposed in MIT Integration Bee Qualifying Exam, 2010, Q24.
Putting both terms into a single fraction reminds one of structure of the derivative of a quotient. Indeed, that is the case:
$$
\int \left(\frac1{\log x} - \frac1{\log^2 x}\right) \diff x
= \int \frac{\log x - 1}{\log^2 x} \diff x
= \int \left(\frac{\diff}{\diff x} \left[\frac{x}{\log x}\right]\right) \diff x
= \color{green}\boxed{\frac{x}{\log x} + C.}
$$
I have selected the last two from this year (2026). The first one only requires a couple trigonometric identities:
Find the following antiderivative
$$
\int \sqrt{\frac{\cos x \cot x \csc x}{\sin x \tan x \sec x}} \diff x.
$$
Proposed in MIT Integration Bee Qualifying Exam, 2026, Q16.
Notice that the integrand simplifies into $\cot^2 x$, so
$$
\int \sqrt{\frac{\cos x \cot x \csc x}{\sin x \tan x \sec x}} \diff x
= \int \cot^2 x \diff x = -x + \int \frac{\diff x}{\sin^2 x} = \color{green}\boxed{-x - \cot x + C.}
$$
Finally, this last one looks funny, but it takes more time to typeset than to solve it:
Evaluate the integral
$$
{\rm I} = \lint_0^{1000} \left(
\lfloor\lceil x\rceil\rfloor + \lceil\lfloor x\rfloor\rceil + \lfloor\{x\}\rfloor
+ \{\lfloor x\rfloor\} + \lceil\{x\}\rceil + \{\lceil x\rceil\}\right) \diff x.
$$
Proposed in MIT Integration Bee Qualifying Exam, 2026, Q15.
Notice that from the definition of the floor, ceil, and fractionary part functions, the third, fourth, and sixth summands are zero; the fifth one is always 1,
and the first two are simply the floor and the ceil respectively (since these operations acting on integers are idempotent). Hence
$$
{\rm I} = 1000 + \lint_0^{1000} \lfloor x\rfloor \diff x + \lint_0^{1000} \lceil x\rceil \diff x
= 1000 + \sum_{k = 0}^{999} k + \sum_{k = 1}^{1000} k = 2000 + 999\cdot 1000 = \color{green}\boxed{1001000.}
$$
All of these (and more) are available at the MIT Integration Bee website.