Computing the sixth derivative of $F(x) = \int_1^x\sin^3(1-t)\mathrm dt$ 
Compute the sixth derivative at $x_0 = 1$ of
  $$F(x) = \int_1^x\sin^3(1-t)\mathrm dt$$

It's from a multiple choice test. I was able to narrow down the choices to $0$ and $60$. I guessed $0$ and obviously it was the other one.
How do I compute that? The first derivative is
$$\sin^3(1 - x)$$
but after the third one it gets hellish. Is there a fast way to compute it? Maybe using some series expansion trick?
 A: As an alternative approach (avoiding trig identities you might not remember), given that you found the first derivative to be $\sin^3(1-x)$, let us write $t=1-x$ for now and recall the Taylor expansion:
$$ \sin^3(t) = \left[t-\frac{t^3}{6}+\frac{t^5}{120} + O(t^7)\right]^3. $$
It is easy enough to show via dumb, direct algebra, that
$$ \sin^3(t) = \left[t-\frac{t^3}{6}+\frac{t^5}{120} + O(t^7)\right]^3 = t^3-\frac{t^5}{2}+O(t^7). $$
Now, we take 5 derivatives, substitute $t=0$ (since that corresponds to $x=1$), and take care to multiply by $(-1)^5$, since $\partial_t=-\partial_x$:
$$ \frac{d^5}{dx^5}\sin^3(1-x)\big|_{x=1} = (-1)^5 \frac{d^5}{dt^5} \sin^3(t)\big|_{t=0} = -(-1)^5 \frac{5!}{2} + O(t^2)\big|_{t=0} = 60.$$
A: You could use the trig identity
$$\cos 3u=4\cos^3 u-3\cos u$$
which leads to
$$\cos^3 u=\frac 14(\cos 3u + 3\cos u)$$
Replacing $u$ with its complement $\frac{\pi}2-u$ we get
$$\cos^3\left(\frac{\pi}2-u\right)=\frac 14\left[\cos\left(\frac{3\pi}2-3u\right) + 3\cos\left(\frac{\pi}2-u\right)\right]$$
$$\sin^3 u=\frac 14(-\sin 3u + 3\sin u)$$
Substitute $u=1-x$ and this will be much easier to differentiate five times.
(I got the initial identity from the third Chebyshev polynomial of the first kind which is fairly well known. The final identity can also be derived by using the sine addition formula twice and replacing $\cos^2 x$ with $1-\sin^2 x$.)
A: $$\begin{eqnarray*} F(x) &=& -\int_{0}^{1-x}\sin^3 t\,dt =\int_{0}^{1-x}\left(\frac{1}{4}\sin(3t)-\frac{3}{4}\sin t\right)\,dt\\&=&-\frac{1}{2}+\frac{3}{4}\cos(x-1)-\frac{1}{12}\cos(3(x-1))\end{eqnarray*}$$
so:
$$ F^{(6)}(x)=-\frac{3}{4}\cos(x-1)+\frac{3^6}{12}\cos(3(x-1)) $$
and:
$$ F^{(6)}(1) = -\frac{3}{4}+\frac{3^5}{4} = \frac{240}{4} = \color{red}{60}.$$
A: the write answser is 60.  In order to compute the derivative you can use the exponential form of the sinus. According to my computation the sixth derivative is 
\begin{equation}
\frac{3}{4}(3^4 cos(3(1-x))- cos(1-x))
\end{equation}
