Dear Uncle Colin,

I need to find the area between the graphs of $y=\sin(2x)$ and $y=\cos^2(x)$ in the interval $0 \le x \lt 2\pi$. I’ve found four solutions, but I think that means I need to do five separate integrals! Is there an easier way?

Trigonometric Expressions: Double Integration Of Ugly Sinusoids

Hi, TEDIOUS, and thanks for your message!

There are several clever ways to find the difference between your curves, other than doing five separate integrals. Let’s check the solutions first:


We can rewrite $\cos^2(x)$ as $\frac{1}{2}\left(\cos(2x) + 1\right)$

That means we need to solve $\sin(2x) = \frac{1}{2}\left(\cos(2x)+1\right)$, or $2\sin(2x) - \cos(2x) = 1$.

We can do a little harmonic substitution there and call the left-hand side $\sqrt{5}\sin(2x - \alpha)$, where $\tan(\alpha) = \frac{2}{\sqrt{5}}$

Our solutions satisfy $\sin(2x - \alpha) = \frac{1}{\sqrt{5}}$, which we can work out numerically as about $0.464$, $\piby2$, $3.605$ and $\frac{3}{2}\pi$. (I’m going to call $\beta \approx 0.464$ to save on writing later.


Take the graph that you’ve clearly sketched, take the bit to the left of $\piby 2$ and move that to the other end so it ends up at $\frac{5}{2}\pi$. Now there are only four regions, and we haven’t changed the area at all.

Also, the area between $\piby2$ and $\pi + \beta$ is the same as that between $\frac{3}{2}\pi$ and $2\pi + \beta$; similarly, the two smaller areas are the same.

So, if we let $f(x) = \frac{1}{2}(2\sin(2x) - \cos(2x) - 1)$, we need to work out $I = \int_{\piby 2}^{\pi+\beta} -f(x) \dx + \int_{\pi+\beta}^{\frac{3}{2}\pi}f(x)\dx$ and double the result.

The indefinite integral of $f(x)$ is $F(x) = -\frac{1}{2}\left(\cos(2x) + \frac{1}{2}\sin(2x) + x\right) + C$, so our final area is $2\left( \left[ F(x) \right]_{\piby2}^{\pi+\beta} + \left[ F(x) \right]_{\pi+\beta}^{\frac{3}{2}\pi}\right)$.

We get something horrible involving $\sqrt{5}$s and $\arctan\left(\frac{1}{2}\right)$s that works out numerically as about 4.9273.


The Wolf tells me the exact answer is $2\left(2 + \arctan\left(\frac{1}{2}\right)\right)$, which suggests there’s a nicer method still! However, I don’t see it. Maybe one of the readers will!

Hope that helps,

- Uncle Colin

* Edited 2020-07-02 to correct a sign error. Thanks to @over_drawn for pointing it out!