Rules of calculus

By the sine addition formula, the integrand becomes

$\large I = \int_{0}^{\frac{\pi}{2}} \sin(2017x) \cos(x) \sin^{2016}(x) + \cos(2017x) \sin^{2017}(x) dx$

By observation, we can use the inverse product rule on this to get

$\large I = \dfrac{\sin(2017x) \sin^{2017} (x)}{2017}$ , so upon evaluating the limits we get $I = \frac{1}{2017}$ .

Here's another solution. While I think it's interesting in its own right, it's much more complicated than the others' solutions, so tread carefully!

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First, we use $\sin(u) = \frac{e^{ui}-e^{-ui}}{2i}$ along with the binomial theorem to expand the integrand $\begin{aligned}\sin(2018x)\left(\sin x\right)^{2016} &= \left(\frac{e^{2018xi}-e^{-2018xi}}{2i}\right)\left(\frac{e^{xi}-e^{-xi}}{2i}\right)^{2016} \\ &= \frac{1}{2^{2016}\cdot(2i)} \left(e^{2018xi}\left(e^{xi}-e^{-xi}\right)^{2016} - e^{-2018xi}\left(e^{xi}-e^{-xi}\right)^{2016}\right) \\ &= \frac{1}{2^{2016}\cdot(2i)} \left(\left(e^{2018xi}\sum_{k=0}^{2016}\binom{2016}{k}\left(e^{xi}\right)^k\left(-e^{-xi}\right)^{2016-k}\right) - \left(e^{-2018xi}\sum_{k=0}^{2016}\binom{2016}{k}\left(e^{xi}\right)^{2016-k}\left(-e^{-xi}\right)^k\right)\right) \\ &= \frac{1}{2^{2016}} \sum_{k=0}^{2016} \binom{2016}{k} (-1)^k \frac{e^{(2k+2)xi}-e^{-(2k+2)xi}}{2i} \\ &= \frac{1}{2^{2016}} \sum_{k=0}^{2016} \binom{2016}{k} (-1)^k \sin((2k+2)x) \end{aligned}$

Now we integrate from $x=0$ to $x=\pi/2$ to find $\begin{aligned} I &= \frac{1}{2^{2016}} \sum_{k=0}^{2016} \binom{2016}{k} (-1)^k \int_0^{\pi/2} \sin((2k+2)x) \, dx \\ &= \frac{1}{2^{2016}} \sum_{k=0}^{2016} \binom{2016}{k} (-1)^k \left(\frac{1}{2k+2}(-\cos((k+1)\pi) + \cos(0))\right) \\ &= \frac{1}{2^{2016}} \sum_{k=0}^{2016} \binom{2016}{k} (-1)^k \cdot \frac{1 - (-1)^{k+1}}{2k+2} \\ &= \frac{1}{2^{2016}} \sum_{k=0}^{1008} \binom{2016}{2k} \frac{1}{2k+1}\end{aligned}$

Now, to evaluate this last expression, we consider the following identity from the binomial theorem: $(1+x)^{2016} = \sum_{k=0}^{2016}\binom{2016}{k} x^k.$ Now, that's similar but not the same as the sum we want. We only want even values of $k$ in the binomial coefficient, which means we only want to take the terms where $x$ has an even exponent. To do this, we take the even part: $\frac{1}{2}\left((1+x)^{2016} + (1-x)^{2016}\right) = \sum_{k=0}^{1008} \binom{2016}{2k} x^{2k}$ Finally, since $\int_0^1 x^{2k}\,dx = \frac{1}{2k+1}$ , apply this integral to both sides to find $\frac{1}{2(2017)}\left((1+x)^{2017} - (1-x)^{2017}\right|_0^1 = \sum_{k=0}^{1008} \binom{2016}{2k}\frac{1}{2k+1} \\ \frac{2^{2016}}{2017} = \sum_{k=0}^{1008} \binom{2016}{2k}\frac{1}{2k+1}$

Using this in the equation for $I$ , we finally find $I = \frac{1}{2^{2016}} \cdot \frac{2^{2016}}{2017} = \frac{1}{2017}$ and therefore $\boxed{I^{-1} = 2017}$

<phew!>

Similar solution with @Andre Bourque 's

$\begin{aligned} I & = \int_0^\frac \pi 2 {\color{#3D99F6}\sin(2018x)}\sin^{2016}x \ dx & \small \color{#3D99F6} \text{By }\sin (A+B) = \sin A \cos B + \cos A \sin B \\ & = \int_0^\frac \pi 2 {\color{#3D99F6}(\sin(2017x)\cos x +\cos (2017x)\sin x)}\sin^{2016}x \ dx \\ & = \int_0^\frac \pi 2 \left({\color{#3D99F6}\sin(2017x)\cos x\sin^{2016}x +\cos (2017x)\sin^{2017}x}\right) \ dx & \small \color{#3D99F6} \text{See note} \\ & = \frac {\sin(2017x)\sin^{2017}x}{2017}\bigg|_0^\frac \pi 2 \\ & = \frac 1{2017} \end{aligned}$

Therefore, $I^{-1} = \boxed{2017}$ .

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Note: $\dfrac {d}{dx} \sin(2017x) \sin^{2017}x = 2017\cos(2017x) \sin^{2017}x + 2017\sin(2017x) \sin^{2016}\cos x$