$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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With $\ds{a > - 1}$ and $\ds{k \in \mathbb{Z}}$:
Note that
\begin{align}
\mrm{I}_{k}\pars{\alpha} & \equiv
\int_{0}^{\pi}\bracks{\sin\pars{t}}^{\,\alpha}\cos\pars{kt}\,\dd t
\\[5mm] & =
\int_{0}^{\pi/2}\bracks{\sin\pars{t}}^{\,\alpha}\cos\pars{\verts{k}t}\,\dd t +
\int_{\pi/2}^{\pi}\bracks{\sin\pars{t}}^{\,\alpha}\cos\pars{\verts{k}t}\,\dd t
\\[5mm] = &\
\int_{0}^{\pi/2}\bracks{\sin\pars{t}}^{\,\alpha}\cos\pars{\verts{k}t}\,\dd t +
\int_{-\pi/2}^{0}\bracks{-\sin\pars{t}}^{\,\alpha}\pars{-1}^{\verts{k}}\cos\pars{\verts{k}t}\,\dd t
\\[5mm] = &\
\bracks{1 + \pars{-1}^{\verts{k}}}
\int_{0}^{\pi/2}\bracks{\sin\pars{t}}^{\,\alpha}\cos\pars{\verts{k}t}\,\dd t
\implies
\bbx{\ds{\mrm{I}_{k}\pars{\alpha} = 0\,,\quad k\ \mbox{odd}}}
\end{align}
Then, for $\ds{\verts{k} = 2m}$
even
$\ds{\pars{m \in \mathbb{Z}_{\geq 0}}}$:
\begin{align}
\left.\mrm{I}_{k}\pars{\alpha}\right\vert_{\ k\ even} & =
2\int_{0}^{\pi/2}\bracks{\sin\pars{t}}^{\,\alpha}\cos\pars{2mt}\,\dd t =
2\,\Re\int_{0}^{\pi/2}\bracks{\sin\pars{t}}^{\,\alpha}\exp\pars{2mt\,\ic}\,\dd t \\[5mm] & =
\left.2\,\Re\int_{0}^{\pi/2}\pars{z - 1/z \over 2\ic}^{\alpha}\,z^{2m}
\,{\dd z \over \ic z}\,
\right\vert_{\ z\ =\ \exp\pars{\ic t}}
\\[5mm] & =
\left.2^{1 - \alpha}\,\Im\int_{0}^{\pi/2}
\pars{{1 - z^{2} \over z}\,\ic}^{\alpha}z^{2m - 1}\,\dd z
\,\right\vert_{\ z\ =\ \exp\pars{\ic t}}\label{1}\tag{1}
\end{align}
I'll 'close' the integral along $\ds{y \in \pars{0,1}}$ and
$\ds{x \in \pars{0,1}}$ and along an indent around $\ds{z = 0}$. The $\ds{\xi^{\alpha}}$ $\mbox{branch-cut}$ is given by
$$
\xi^{\alpha} = \verts{\xi}^{\alpha}\exp\pars{a\,\mrm{arg}\pars{\xi}\ic}\,;
\qquad -\pi < \mrm{arg}\pars{\xi} < \pi\,,\quad \xi \not= 0
$$
\eqref{1} becomes
\begin{align}
\left.\mrm{I}_{k}\pars{\alpha}\right\vert_{\ k\ even} & =
-2^{1 - \alpha}\,\Im\int_{1}^{\epsilon}\pars{1 + y^{2} \over y}^{\alpha}\,
\pars{-1}^{m + 1}\,\ic\,y^{2m - 1}\ic\,\dd y
\\[5mm] & -
2^{1 - \alpha}\,\Im\int_{\pi/2}^{0}
\bracks{{1 \over \epsilon\exp\pars{\ic t}}\,\ic}^{\alpha}\,
\epsilon^{2m - 1}\exp\pars{\ic\bracks{2m - 1}t}\,
\epsilon\exp\pars{\ic t}\ic\,\dd t
\\[5mm] & -
2^{1 - \alpha}\,\Im\int_{\epsilon}^{1}\pars{1 - x^{2} \over x}^{\alpha}
\exp\pars{\ic\,{\pi \over 2}\,\alpha}\,x^{2m - 1}\,\dd x
\\[1cm] & =
2^{1 - \alpha}\epsilon^{2m - \alpha}\,\,\,\,
\overbrace{\Re\int_{0}^{\pi/2}
\exp\pars{\ic\bracks{{\pi \over 2} - t}\alpha}\exp\pars{2mt\,\ic}\,\dd t}
^{\ds{-\,{\sin\pars{\pi\alpha/2} \over 2m - \alpha}}}
\\[5mm] & -
2^{1 - \alpha}\sin\pars{{\pi \over 2}\,\alpha}
\int_{x\ =\ \epsilon}^{x\ =\ 1}\pars{1 - x^{2}}^{\alpha}
\,\dd\pars{x^{2m - \alpha} \over 2m - \alpha}
\end{align}
Integrating by parts the last integral and taking the limit $\ds{\epsilon \to 0^{+}}$:
\begin{align}
\left.\mrm{I}_{k}\pars{\alpha}\right\vert_{\ k\ even} & =
{2^{1 - \alpha}\sin\pars{\pi\alpha/2} \over 2m - \alpha}
\int_{0}^{1}x^{2m - \alpha}\,\alpha\pars{1 - x^{2}}^{\alpha - 1}\pars{-2x}
\,\dd x
\\[5mm] & \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{2^{1 - \alpha}\alpha\sin\pars{\pi\alpha/2} \over \alpha - 2m}
\int_{0}^{1}x^{m - \alpha/2}\pars{1 - x}^{\alpha - 1}\,\dd x
\\[5mm] & =
{2^{1 - \alpha}\alpha\sin\pars{\pi\alpha/2} \over \alpha - 2m}\,
{\Gamma\pars{m - \alpha/2 + 1}\Gamma\pars{\alpha} \over
\Gamma\pars{m + \alpha/2 + 1}}
\\[5mm] & =
{2^{1 - \alpha}\sin\pars{\pi\alpha/2} \over \alpha - 2m}\,
{\bracks{\pars{m - \alpha/2}\Gamma\pars{m - \alpha/2}}
\bracks{\alpha\Gamma\pars{\alpha}} \over \Gamma\pars{m + \alpha/2 + 1}}
\\[5mm] & =
-2^{-\alpha}\sin\pars{\pi\alpha/2}
{\Gamma\pars{m - \alpha/2} \Gamma\pars{\alpha + 1} \over
\Gamma\pars{m + \alpha/2 + 1}}
\\[5mm] & =
-2^{-\alpha}\sin\pars{\pi\alpha/2}\,
{\pi \over \Gamma\pars{1 - m + \alpha/2}\sin\pars{\pi m - \pi\alpha/2}}
{\Gamma\pars{\alpha + 1} \over \Gamma\pars{m + \alpha/2 + 1}}
\\[5mm] & =
2^{-\alpha}\pi\,{\Gamma\pars{\alpha + 1} \over
\Gamma\pars{1 - m + \alpha/2}\Gamma\pars{m + \alpha/2 + 1}}\,\cos\pars{\pi m}
\end{align}
Finally:
$$\bbx{\ds{%
\int_{0}^{\pi}\bracks{\sin\pars{t}}^{\,\alpha}\cos\pars{kt}\,\dd t
=
{\pi\,2^{-\alpha}\,\Gamma\pars{\alpha + 1}\cos\pars{\pi k/2} \over
\Gamma\pars{\bracks{2 - k + \alpha}/2}\Gamma\pars{\bracks{k + \alpha + 2}/2}}}}
$$
