TITLE: Calculating $\int_0^{\pi/2} (x \sin(x))^n dx$
QUESTION [6 upvotes]: Define
$$I_n = \int_0^{\pi/2} (x \sin(x))^n dx$$
for $n \ge 0$.
I calculate the value for $n = 0, 1$ and $2$.
$$I_0 = \frac{\pi}{2}, I_1 = 1, I_2 = \frac{{\pi}^3 + 6 \pi}{48} .$$
In general, what's the value of $I_n$?
P.S.
By WolframAlpha(
n = 3 (http://www.wolframalpha.com/input/?i=integrate+%5B(x+sin(x))%5E3,+%7Bx,+0,+PI%2F2%7D%5D),
n = 4 (http://www.wolframalpha.com/input/?i=integrate+%5B(x+sin(x))%5E4,+%7Bx,+0,+PI%2F2%7D%5D),
n = 5 (http://www.wolframalpha.com/input/?i=integrate+%5B(x+sin(x))%5E5,+%7Bx,+0,+PI%2F2%7D%5D)
),
$$I_3 = \frac{7{\pi}^2}{12} - \frac{122}{27},$$ 
$$ I_4 = \frac{6{\pi}^5 + 170 {\pi}^3 -975 \pi}{2560},$$
$$ I_5 = \frac{149{\pi}^4}{720} - \frac{31841{\pi}^2}{3375} + \frac{56992552}{759375}.$$

REPLY [2 votes]: Possible hint, or end of the story.
$$\int_0^{\pi/2} x^n\sin^n(x)\ \text{d}x = \int_0^{\pi/2} x^n\ \left(\frac{e^{ix} - e^{-ix}}{2i}\right)^n\ \text{d}x$$
But I warn you, it's not an easy road. You'll have to integrate by parts $n$ times. You'll end up almost surely with some hellish Hypergeometric Function.
Synthesis: no clue if there really is a general close form.
P.s. Binomial expansion may be useful.