Evaluating $\int_0^\pi \sin^2(3x+\cos^45x)dx$ (2023 MIT Integration Bee #16)

Question

2023 MIT Integration Bee Regular Season, Problem 16. $$\int_0^\pi \sin^2(3x+\cos^45x)dx$$

I got as far as $$ \int_0^\pi \sin^2(3x+\cos^45x)dx = \frac{1}{2}\int_0^\pi 1dx \\ - \frac{1}{2}\int_0^\pi \cos(6x+2\cos^45x)dx $$ For the 2nd integral, I’ve tried substitutions like $x\rightarrow\pi-x$ and $x\rightarrow\frac{\pi}{2}-x$ but they don’t seem to lead anywhere. Would appreciate some help in how to show that the 2nd integral is $0$ so that the final answer is $\frac{\pi}{2}$, or if there’s another method completely to go about this.

Answer 1

These solutions are all very well thought out, but an integration bee problem has to be solved in minutes. Most likely the intent behind the problem was to recognize that

$$\int_0^\pi \sin^2(3x+\cos^4(5x))\:dx = \int_0^\pi\sin^2(3x)\:dx$$

or in other words the cosine term doesn't matter, which solves the problem immediately. But how would one quickly justify that to themselves?

Consider

$$I[a] = \int_0^\pi\sin^2(3x+a\cos^4(5x))\:dx \implies I'[a] = \int_0^\pi\sin(6x+2a\cos^4(5x))\cos^4(5x)\:dx$$

There are a few immediate simplifications to be made here. First using the sine angle addition formula

$$\require{cancel} I'[a] = \cancelto{0}{\int_0^\pi\sin(6x)\cos(2a\cos^4(5x))\cos^4(5x)\:dx}+\int_0^\pi\cos(6x)\sin(2a\cos^4(5x))\cos^4(5x)\:dx$$

which cancels due to odd symmetry on the interval. Next, observe that the remaining integrand is $\pi$-periodic, so we can move the integral over any interval of $\pi$ length we wish. Choosing $[0,\pi]\to\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ we get

$$I'[a] = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(6x)\sin(2a\cos^4(5x))\cos^4(5x)\:dx = \cancelto{0}{2\int_0^{\frac{\pi}{2}}\cos(6x)\sin(2a\cos^4(5x))\cos^4(5x)\:dx}$$

due to even then odd symmetry. Thus $I[a]$ is a constant so

$$I[a] = I[0] = \int_0^\pi\sin^2(3x)\:dx = \frac{\pi}{2}$$

Explaining the steps out makes this seem like a long process, but on a bee contestants would not have to think the symmetry computations all the way through to justify to themselves to a good degree of confidence that things cancel out. The same cannot be said for computations involving series.

Answer 2

Let $$I = \int_{0}^{\pi}\cos{(6x + 2\cos^{4}{5x})} \ dx$$ Let $x = \pi - u$ $$I = \int_{0}^{\pi}\cos{(6(\pi - u) + 2\cos^{4}{5(\pi - u)})} \ du = \int_{0}^{\pi}\cos{(-6u + 2\cos^{4}{5u})} \ dx$$ $$=\int_{0}^{\pi}\cos{(-6x + 2\cos^{4}{5x})} \ dx$$ Thus, $$2I = \int_{0}^{\pi}\cos{(6x + 2\cos^{4}{5x})} + \cos{(-6x + 2\cos^{4}{5x})} \ dx$$ $$= \int_{0}^{\pi}2\cos{(6x)}\cos{(2\cos^{4}{(5x)})} \ dx$$ Hence, $$I = \int_{0}^{\pi}\cos{(6x)}\cos{(2\cos^{4}{(5x)})} \ dx$$ We rewrite using the Maclaurin series of $\cos{x}$, $$I = \int_{0}^{\pi}\cos{(6x)}\sum_{n=0}^{\infty}\frac{(-1)^{n}2^{n}\cos^{8n}(5x)}{(2n)!} \ dx = \sum_{n=0}^{\infty}(-1)^{n}\frac{2^{n}}{(2n)!}\int_{0}^{\pi}\cos{(6x)}\cos^{8n}(5x) \ dx$$ Using orthogonality relations and trig identities, you can show that for each $n\in \mathbb{N}$, $$\int_{0}^{\pi}\cos(6x)\cos^{8n}(5x) \ dx = 0$$ Hence, $$I = 0$$

Answer 3

$$ \begin{aligned} \int_0^\pi \sin ^2\left(3 x+2 \cos ^4 5 x\right) d x = & \frac{1}{2} \int_0^\pi\left[1-\cos \left(6 x+4 \cos ^4 5 x\right)\right] d x \\ = & \frac{\pi}{2}-\frac{1}{2} \underbrace{\int_0^\pi \cos \left(6 x+4 \cos ^4 5 x\right) d x}_{J} \end{aligned} $$ Noting that $x\mapsto \pi-x$ gives $$J= \int_0^\pi \cos \left(4 \cos ^4 5 x-6 x\right) d x$$ Averaging them yields $$ \begin{aligned} J & =\frac{1}{2}\int_0^\pi \left[\cos \left(4 \cos ^4 5 x+6 x\right)+\cos \left(4 \cos ^4 5 x-6 x\right) \right]d x \\ & =\int_0^\pi \cos \left(4 \cos ^4 5 x\right) \cos 6 xd x\\&= \Re\int_0^\pi e^{4 i \cos ^4 5 x} \cos 6 x d x \end{aligned} $$ Expanding by definition of $e^x$ yields $$ J=\Re \sum_{n=0}^{\infty} \frac{(4 i)^n}{n !} \int_0^\pi \cos ^{4 n} 5 x \cos 6 x d x $$ For the last integral, we express the integrand in terms of $\cos mx\cos 5x$ as below:

$$ \begin{aligned} \cos ^{4 n} 5 x&=\left(\frac{e^{5 x i}+e^{-5 x i}}{2}\right)^{4 n} \\ & =\frac{1}{2^{4 n}} \sum_{k=0}^{4 n}\left(\begin{array}{c} 4 n \\ k \end{array}\right) e^{5(4 n-2 k) x i} \\ &=\frac{1}{2^{4 n}} \left[\sum_{k=0}^{2 n-1}\left(\begin{array}{c} 4 n \\ k \end{array}\right) \cos [5(4 n-2 k) x]+\left(\begin{array}{c} 4 n \\ 2 n \end{array}\right)\right] \\ & \end{aligned} $$ Hence $$ \int_0^\pi \cos ^{4 n} 5 x \cos 6 x d x= \frac{1}{2^{4 n}} \left[\sum_{k=0}^{2 n-1}\left(\begin{array}{c} 4 n \\ k \end{array}\right)\int_0^\pi \cos [5(4 n-2 k) x]\cos 6x dx + \left(\begin{array}{c} 4 n \\ 2 n \end{array}\right) \int_0^\pi\cos 6xdx\right] =0 $$ Hence we can conclude that $$\boxed{\int_0^\pi \sin ^2\left(3 x+2 \cos ^4 5 x\right) d x = \frac{\pi}{2}}$$

Answer 4

$$ \begin{align} I &:= \int_{0}^{\pi}\sin^{2}\left(3x+2\cos\left(5x\right)^{4}\right)dx \\ &= \frac{1}{2}\Re\int_{0}^{\pi}\left(1-\exp\left(2i\left(3x+2\cos^{4}\left(5x\right)\right)\right)\right)dx \\ &= \frac{1}{2}\int_{0}^{\pi}dx-\frac{1}{2}\Re\int_{0}^{\pi}\exp\left(6ix+4i\cdot\frac{1}{4}\left(1+2\cos\left(10x\right)+\frac{1}{2}\left(1+\cos\left(20x\right)\right)\right)\right)dx \\ &= \frac{\pi}{2}-\frac{1}{2}\Re\exp\left(i+\frac{i}{2}\right)\int_{0}^{\pi}e^{6ix}\exp\left(2i\cos\left(10x\right)\right)\exp\left(\frac{i}{2}\cos\left(20x\right)\right)dx \\ &= \frac{\pi}{2}-\frac{1}{2}\Re\exp\left(i+\frac{i}{2}\right)\int_{0}^{\pi}e^{6ix}\sum_{n=-\infty}^{\infty}i^{n}\operatorname{J}_{n}\left(2\right)e^{10nix}\sum_{k=-\infty}^{\infty}i^{k}\operatorname{J}_{k}\left(\frac{1}{2}\right)e^{20kix}dx \tag{1}\\ &= \frac{\pi}{2}-\frac{1}{2}\Re\exp\left(i+\frac{i}{2}\right)\sum_{n,k=-\infty}^{\infty}i^{n+k}\operatorname{J}_{n}\left(2\right)\operatorname{J}_{k}\left(\frac{1}{2}\right)\int_{0}^{\pi}e^{6ix}e^{10nix}e^{20kix}dx \\ &= \frac{\pi}{2}-\frac{1}{2}\Re\exp\left(i+\frac{i}{2}\right)\sum_{n,k=-\infty}^{\infty}i^{n+k}J_{n}\left(2\right)J_{k}\left(\frac{1}{2}\right)\int_{0}^{\pi}e^{6ix}e^{10nix}e^{20kix}dx \\ &= \frac{\pi}{2}-\frac{1}{2}\Re\exp\left(i+\frac{i}{2}\right)\sum_{n,k=-\infty}^{\infty}i^{n+k}\operatorname{J}_{n}\left(2\right)\operatorname{J}_{k}\left(\frac{1}{2}\right)\cdot\frac{i\left(1-e^{2\pi i\left(10k+5n+3\right)}\right)}{20k+10n+6} \\ &= \frac{\pi}{2}-\frac{1}{2}\Re\exp\left(i+\frac{i}{2}\right)\sum_{n,k=-\infty}^{\infty}i^{n+k}\operatorname{J}_{n}\left(2\right)\operatorname{J}_{k}\left(\frac{1}{2}\right)\cdot 0 \\ &= \frac{\pi}{2} \end{align} $$

Where in $(1)$ we used the Jacobi–Anger expansion.