Evaluate $\int_0^{\pi/2}(\sin x)^n e^{-(2+\cos x)\log k}dx$ for fixed integers $n,k\geq 1$ My question is the following

Question. Can you compute some of the following $$c_{n,k}=\int_0^{\pi/2}(\sin x)^n e^{-(2+\cos x)\log k}dx$$
where $n\geq 1$ is a fixed integer  and $k\geq 1$ is also a fixed integer? What of those? Thanks in advance.

If you know some of related integrals you can add also a link from this site of those computations that provide closed forms for our integrals in previous Question. For the other, you can add yourself computations/reductions.
My attempt was convice to me that seems a hard problem, because for the lower odds $n=1$ or $n=3$ using the rule $$\int F'(x)e^{F(x)}dx=e^{F(x)}+\text{constant}$$ and integration by part, is a lot of work yet. I take the case $k=1$ as a special case, thus I am knowing that from Wolfram Alpha $\int_0^{\pi/2}(\sin x)^ndx$ a closed form is provide us with the code
$$\int_0^{\pi/2} \sin^n(x)\, dx$$
for integers $n\geq 1$. But the code for a reduction with $n=6$, this is
$$\int_0^{\pi/2} \sin^6(x)\,e^{-\cos(x)}\, dx$$
doesn't provide to me a closed-form. For example $n=5$ likes to me tedious computations.
The context of my computations is speculative, I don't know if there were mistakes or my reasoning/approach is reasonable (please, feel free to add comments if you detect some): let the complex variable $s=\sigma+it$ , then $\zeta(s)$ the Riemann Zeta function is analytic for $\Re s>1$ and doen't vanishes. Is known that $$\frac{1}{\zeta(s)}=\sum_{n=1}^\infty\frac{\mu(n)}{n^s},$$
for $\Re s>1$. I believe that I can define the following real functions, as restrictions ($X=x+0\cdot i$) of previous complex function,  $$f(X)=\frac{1}{\zeta(X+2)}$$
as continuous for $-1<X\leq 1$, thus this next composition $f: \left[ 0,1\right] \to\mathbb{R}$,  also as continuous real function $$f(\cos x)=\frac{1}{\zeta(2+\cos x)}=\sum_{k=1}^\infty\mu(k)k^{-2-\cos x},$$
when (if I can presume it) $$-1<\cos x\leq 1.$$
Then as a specialization of the PROBLEMA 159 in page 515, in spanish, proposed by Furdui in La Gaceta de la RSME Vol. 14 (2011) No. 3, I say the first statement, and the first line of the Solution that provide us
$$\frac{\sqrt{\frac{\pi}{2}}}{\zeta(3)}=\lim _{n\to\infty}\sqrt{n}\sum_{k=1}^\infty\mu(k)c_{n,k},$$
when is justified to swap the sign of  series $\sum_{k=1}^\infty$ with the definite integral.
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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\begin{align}
\color{#f00}{c_{n,k}} & \equiv
\int_{0}^{\pi/2}\sin^{n}\pars{x}\expo{-\bracks{2 + \cos\pars{x}}\ln\pars{k}}
\,\dd x =
{1 \over k^{2}}\int_0^{\pi/2}\sin^{n}\pars{x}\,k^{-\cos\pars{x}}\,\dd x
\\[3mm] & =
-\,{1 \over k^{2}}\int_{x = 0}^{x = \pi/2}
\bracks{1 - \cos^{2}\pars{x}}^{\pars{n - 1}/2}\,
k^{-\cos\pars{x}}\,\dd\bracks{\cos\pars{x}}
\\[3mm] & \stackrel{\cos\pars{x}\ \to\ x}{=}\
{1 \over k^{2}}\int_{0}^{1}\pars{1 - x^{2}}^{\pars{n - 1}/2}\,k^{-x}\,\dd x =
{1 \over k^{2}}\int_{0}^{1}\pars{1 - x^{2}}^{\pars{n - 1}/2}\,\expo{-\mu x}
\,\dd x
\\[3mm] &\ \mbox{where}\ \mu \equiv \ln\pars{k}
\end{align}

Then,
\begin{align}
\color{#f00}{c_{n,k}} & =
{1 \over k^{2}}\bracks{%
-\,{\root{\pi}\Gamma\pars{n/2 + 1/2} \over 2\pars{\mu/2}^{n/2}}\,
\mathbf{\mathrm{M}}_{n/2}\pars{\mu}}
\\[3mm] & =
\color{#f00}{-\,{\root{\pi} \over k^{2}}2^{n/2 - 1}\ln^{-n/2}\pars{k}\,
\Gamma\pars{{n \over 2} + \half}\mathbf{\mathrm{M}}_{n/2}\pars{\ln\pars{k}}}\,,
\qquad \Re\pars{n} > -1
\end{align}

$\mathbf{\mathrm{M}}_{\nu}\pars{z}$ is the Modified Struve Function. See, for example, $\mathbf{11.5.4}$ ( page 292 ) in NIST HandBook of Mathematical Functions or/and DLMF-NIST page.
A: This is not an answer since the result comes from a CAS.
$$\int_{0}^{\frac{\pi}2}\sin^n(x)\,dx=\frac{\sqrt{\pi } \Gamma \left(\frac{n+1}{2}\right)}{2 \Gamma
   \left(\frac{n}{2}+1\right)}$$ $$\int_{0}^{\frac{\pi}2}\sin^n(x)e^{-\cos(x)}\,dx=\sqrt{\pi }\, 2^{\frac{n}{2}-1} \,\Gamma \left(\frac{n+1}{2}\right)
   \left(I_{\frac{n}{2}}(1)-\pmb{L}_{\frac{n}{2}}(1)\right)$$ where appear Bessal and Struve functions. Both are of $\Re(n)>-1$.
It evens seems that $$\int_{0}^{\frac{\pi}2}\sin^n(x)e^{-a\,\cos(x)}\,dx=\sqrt{\pi }\, 2^{\frac{n}{2}-1}\, a^{-n/2}\, \Gamma \left(\frac{n+1}{2}\right)
   \left(I_{\frac{n}{2}}(a)-\pmb{L}_{\frac{n}{2}}(a)\right)$$
May be (I hope), this will give you some ideas.
