Challenging Integral of $1+\sin^2x+\cdots+\sin^{16}x$ 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. 
 A: 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.
A: 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 
$$\frac{8168160\pi-7559999\sqrt3}{7340032}$$
