Non-obvious Trigometric Integral More Context: Integral cosine transforms are of interest in various fields, including image processing, operator theory, and special functions. In particular, the kernel, $\alpha\ge 0$,
$$ F(t) := \sin^{\alpha}(t) \cos^{\alpha-1}(t) (1-(\alpha+2)\cos^2(2t))$$
has appeared to be of interest in my special functions research. I would like to know properties about $\mathscr{T}(F)$, where $\mathscr{T}$ is a consine transform operator. The simplest question might be; where is this transform equal to zero; i.e., can we solve
$$ \mathscr{T}(F) = 0$$
that is, for $z$ and $R$, when is it the case that
$$\int_0^{R} F(t)\cos(zt) = 0 $$
Of surprise to me when doing numerical calculations, was that this appeared to be zero at $R=\frac{\pi}{4}$ and $z=1$, independent of $\alpha$; which was unexpected. Hence, I was motivated to see if any other mathematicians might have some simple insights about whether this is true or not. The question did not appear difficult enough to warrant post on mathoverflow and hence I asked on math.stackexchange.
I could use some explanation about why this context helps answer the original question? What is the motivation behind flagging this question in the first place? Including additional introductory pleasantries appears to detract from the simple question/answer style that I have been expected to adhere to in the past on this platform. But even if it is the case, that policies have changed and that more context is desired; how is that even measured? Including a small paragraph about why I am asking the question, appears to be insufficient? How much context is desired? And to what end? Moreover, including this information does little more than distract from the main question; do we achieve zero at $R=\frac{\pi}{2}$ and $z=1$, independent of $\alpha$?

Context: In researching various integral cosine transforms, I came across this particularly difficult looking problem. Numerical calculations suggest the integral is zero. Is this true?

For every $\alpha\ge 0$ does
$$\int_0^{\pi/4} (\sin(t)\cos(t))^\alpha(1-(\alpha+2)\cos^2(2t)) = 0\ \ ?$$
Not very obvious to me.
 A: Denoting your desired integral by $I$, recalling that $\sin(2x) = 2 \sin(x) \cos(x)$ we get
\begin{align}
 I & =\int_0^{\frac{\pi}{4}} (\sin(t)\cos(t))^\alpha(1-(\alpha+2)\cos^2(2t))\, \mathrm{d}t \\ & \overset{2t\to t}{=} \frac{1}{2^{\alpha+1}}\int_0^{\frac{\pi}{2}} \sin^{\alpha}(t)(1 - (\alpha+2)\cos^2(t))\, \mathrm{d}t\\
&=\frac{1}{2^{\alpha+1}}\int_0^{\frac{\pi}{2}} \sin^{\alpha}(t)(\color{purple}{1-\cos^2(t)} - (\alpha+1)\cos^2(t))\, \mathrm{d}t\\
& = \frac{1}{2^{\alpha+1}}\left[\underbrace{\int_0^{\frac{\pi}{2}} \sin^{\alpha+\color{purple}{2}}(t)\, \mathrm{d}t}_{I_1}- \underbrace{\int_0^{\frac{\pi}{2}}(\alpha+1) \sin^{\alpha}(t)\cos^2(t)\, \mathrm{d}t}_{I_2}\right]
\end{align}
For $I_1$ we do the following:
\begin{align}
I_1 & =\int_0^{\frac{\pi}{2}} \sin^{\alpha+2}(t)\, \mathrm{d}t\\
 &= \int_0^{\frac{\pi}{2}} \sin^{\alpha+1}(t)\underbrace{\left(-\cos(t) \right)'}_{\color{blue}{\sin(t)}}\, \mathrm{d}t\\
& \overset{\text{I.B.P.}}{=} - \sin^{\alpha+1} (t)\cos(t)\Bigg\vert_{0}^{\frac\pi2} + \int_0^{\frac\pi2}(\alpha+1)\sin^{\alpha}(t)\cos^2(t)\, \mathrm{d}t\\
& = I_2
\end{align}
And since $I =\frac{1}{2^{\alpha+1}}\left[I_1 - I_2\right]$ we can conclude that $$\int_0^{\frac{\pi}{4}} (\sin(t)\cos(t))^\alpha(1-(\alpha+2)\cos^2(2t))\, \mathrm{d}t =0$$
