MIT Integration Bee 2023 Semi-final Problem 1

Posted: May 19, 2023 in Mathematics
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I was browsing the net couple days ago and saw this integral. At first it looks quite difficult. I have tried numerous different substitutions and they don’t work. And later I tried using complex number and it seems to work. So, here is my solution.

Solution:

By Euler’s formula, cosine function is the real part of a complex number.

e^{i\theta} = \cos \theta + i \sin \theta

Hence,

\displaystyle \int e^{\cos x} \cos(2x + \sin x) \text{ d}x = \Re \int e^{\cos x} e^{i(2x+\sin x)} \text{ d}x

Re-arrange the exponents we’ll get

\displaystyle = \Re \int e^{\cos x + i\sin x} e^{2ix} \text{ d}x = \Re \int e^{e^{ix}} e^{2ix}\text{ d}x

Now, since

\displaystyle \frac{\text{d}}{\text{d}x} e^{e^{ix}}=ie^{e^{ix}}e^{ix}

We can use integration by parts,

\displaystyle = \Re \; \frac{1}{i} \int e^{ix} \text{ d}e^{e^{ix}}

\displaystyle = \Re \; \frac{1}{i}\left(e^{ix}e^{e^{ix}} - \int e^{e^{ix}}\cdot i e^{ix} \text{ d}x \right)

\displaystyle = \Re \; \frac{1}{i}\left(e^{ix}e^{e^{ix}} - e^{e^{ix}}\right) + C

\displaystyle = \Re \; \frac{1}{i} e^{e^{ix}} \left(e^{ix}-1\right) + C

\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} (\cos x + i\sin x - 1) + C

\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} (\cos x - 1 + i\sin x) + C

By using the double angle identity of sine and cosine,

\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} \left(-2\sin^2 \frac{x}{2} + i2\sin \frac{x}{2} \cos\frac{x}{2}\right) + C

\displaystyle = \Re \; e^{\cos x + i\sin x} \left(2\sin\frac{x}{2}\cos\frac{x}{2} + i\left(2\sin^2 \frac{x}{2}\right)\right) + C

\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x + i\sin x} \left(\cos\frac{x}{2} + i\sin \frac{x}{2}\right) + C

\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x + i\sin x} e^{i\frac{x}{2}} + C

\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x} e^{i\left(\frac{x}{2}+\sin x\right)} + C

\displaystyle = 2 e^{\cos x} \sin\frac{x}{2} \cos\left(\frac{x}{2} + \sin x\right) + C

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