Challenging Integral of $1+\sin^2x+\cdots+\sin^{16}x$

Question

Question

Evaluate the integral

$$ \int_0^{\frac{\pi}{3}} (1+\sin^2x+\cdots+\sin^{16}x) \ dx$$

Attempt

I simplify the GP to $$\frac{1-\sin^{18}x }{ \cos^2x } $$

but at this point, integration seems extremely difficult...

This question appeared in a South Australian Year 12 Examination, so the methods should be elementary.

Wait, there is no $\sin(x)$ term? So powers are $0,2,4,6,8,...,16$? — user202729, Jan 08 '18 at 08:27
Wolfram has the exact form as $$\frac{8168160\pi-7559999\sqrt3}{7340032}$$ — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jan 08 '18 at 08:29
Are you sure the integral is correct as given? WolframAlpha gives an exact answer of the form $\frac{\alpha \pi + \beta \sqrt{3}}{\gamma}$ with $\alpha, \beta, \gamma$ fairly large integers. — anomaly, Jan 08 '18 at 08:30
@anomaly Probably elementary method, but calculated with handheld calculators. — user202729, Jan 08 '18 at 08:30
The general formula of the integral is a polynomial in $x$ and $sin(nx)$ with very large coefficient. — user202729, Jan 08 '18 at 08:32
@user202729 Elementary? And the expanded integrand is only $$\frac1{16384}(-8\cos2x+\cos4x+15)(-1752\cos2x+879\cos4x-284\cos6x+66\cos8x-15\cos10x+\cos12x+3150)$$ — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jan 08 '18 at 08:33
if $I_n=\int \sin^{n}x dx$ then, by integration by parts, $a_nI_n=b_nI_{n-2}+c_n$ where $a_n,b_n,c_n$ are constants easy to compute. — clark, Jan 08 '18 at 08:33
Graphical calculators, that can do numerical integration, are allowed in year 12 in South Australia, so it was probably asking them to find the OP's shortcut. — Empy2, Jan 08 '18 at 08:33
@TheSimpliFire I said "elementary method", not that calculate the integrand is not tedious. Also calculators are allowed, as ^ . — user202729, Jan 08 '18 at 08:35
Are you looking for the exact value or an approximation of the result ? — Claude Leibovici, Jan 08 '18 at 08:44

score 3 · Answer 1 · answered Jan 08 '18 at 08:44

This kind of question can be solved easily using the identity $\sin(x)=\dfrac{e^{ix}+e^{-ix}}{2i}$ and Newton's binomial identity. However, the computations will be kinda ugly in the OP case.

Another way to proceed: set $I_k=\int_0^{\pi/3}\sin^{2k}(x){\rm d}x$.

Then $I_{k+1}-I_k=\int_0^{\pi/3}\sin^{2k}(x)\cos^2(x){\rm d}x=[\frac{1}{2k+1}\sin^{2k+1}(x)\cos(x)]_0^{\pi/3}+\frac{1}{2k+1}I_{2k+1}$.

Then, we find a linear recurrence relation between $I_k$ and $I_{k+1}$, so we may compute $I_0,\ldots,I_8$ and compute the desired integral.

score 3 · Answer 2 · answered Jan 08 '18 at 09:53

I am not sure that this is a serious answer.

If you have a look here (in the section Power-reduction formulae), you will see that, if $n$ is even, $$\sin^n(x) = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} \binom{n}{k} \cos{\big((n-2k)x\big)}$$

This means that, for your case, $32768$ times the integrand is given by $$-112028 \cos (2 x)+49024 \cos (4 x)-17844 \cos (6 x)+5228 \cos (8 x)-1180 \cos (10 x)+$$ $$192 \cos (12 x)-20 \cos (14 x)+\cos (16 x)+109395$$ leading, after integration, to the result already given by TheSimpliFire $$\frac{8168160\pi-7559999\sqrt3}{7340032}$$

Challenging Integral of $1+\sin^2x+\cdots+\sin^{16}x$

2 Answers2