How can we evaluate $\int_{0}^{\pi/2}\sin2x\ln^n(1+\sin x)\mathrm dx?$ How does one go about evaluating $(1)$?

$$\int_{0}^{\pi/2}\sin2x\ln^n(1+\sin x)\mathrm dx\tag1$$
  $n\ge1$

An attempt:
$u=1+\sin x$ $\implies du=\cos x dx$
Then $(1)$ becomes 
$$2\int_{1}^{2}(u-1)\ln^n(u)\mathrm du\tag2$$
$$(2)=I_1+I_2=2\int_{0}^{2}(u-1)\ln^n(u)\mathrm du-2\int_{0}^{1}(u-1)\ln^n(u)\mathrm du\tag3$$
$I_2$ is the gamma function and $I_1$ slightly different of limit, we don't how to Evaluate it.
Can someone give us a full evaluation of $(1)$, please. Thank you.
 A: 
$$2\int_{1}^{2}(u-1)\ln^n(u)\mathrm du\tag2$$

We now use the substitution $u=e^z$ and $du=e^z dz$.
Continuing from $(2)$, we get
$$2\int_{0}^{\ln 2}(e^z-1)z^n \cdot e^z\mathrm dz$$
$$=2\int_{0}^{\ln 2}z^n e^{2z} \mathrm dz - 2\int_{0}^{\ln 2} z^n e^z \mathrm dz$$
$$=\frac{1}{2^n}\int_{0}^{\ln 2}(2z)^n e^{2z} \mathrm d(2z) - 2\int_{0}^{\ln 2} z^n e^z \mathrm dz$$
Putting $p=2z$, we now have
$$\frac{1}{2^n}\int_{0}^{2\ln 2}p^n e^p \mathrm dp - 2\int_{0}^{\ln 2} z^n e^z \mathrm dz$$
Now integrate by parts $n$ times. See what you get.
For the integral $$\int_{0}^{c} s^n e^s \mathrm ds$$,
We finally get $$[c^n-nc^{n-1}+n(n-1)c^{n-2}- \ldots +(-1)^n n!]e^c - (-1)^n n!$$
A: Here's what I get...
$$\int\sin (2 x) \log ^n(\sin (x)+1) dx $$
$$= 2^{-n} (-\log (\sin (x)+1))^{-n} \log ^n(\sin (x)+1) \Gamma (n+1,-2 \log (\sin (x)+1))-2 (-\log (\sin (x)+1))^{-n} \log ^n(\sin (x)+1) \Gamma (n+1,-\log (\sin (x)+1))\biggr\rvert^{\pi/2}_0$$
$$=(-2)^{-n} \Gamma (n+1,-2 \log (2))-2 (-1)^{-n} \Gamma (n+1,-\log (2)) - 2^{-n} \Gamma (n+1,0)-2 \Gamma (n+1,0)$$
$$ = \left(2-2^{-n}\right) \Gamma (n+1,0)+(-2)^{-n} \left(\Gamma (n+1,-\log (4))-2^{n+1} \Gamma (n+1,-\log (2))\right)$$
A: $$J(n)=\int_{1}^{2}(u-1)\log^n(u)\,du = \left.\frac{d^n}{d\alpha^n}\int_{1}^{2}(u-1)u^{\alpha}\,du\,\right|_{\alpha=0} \tag{4}$$
hence:
$$ J(n) = \left.\frac{d^n}{d\alpha^n}\left(\frac{2^{\alpha+2}-1}{\alpha+2}-\frac{2^{\alpha+1}-1}{\alpha+1}\right)\right|_{\alpha=0}.\tag{5} $$
