Let $I=\displaystyle\int_{0}^{\pi}x^{3}\sin x\,dx$.
Using integration by parts: let $u=x^{3},\; dv=\sin x\,dx$ $\Rightarrow du=3x^{2}dx,\; v=-\cos x$
$I = [-x^{3}\cos x]_0^{\pi} + \displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx$
Now, let $J=\displaystyle\int_{0}^{\pi}x^{2}\cos x\,dx$
Again, by parts: $u=x^{2},\; dv=\cos x\,dx \Rightarrow du=2x\,dx,\; v=\sin x$
$J = [x^{2}\sin x]_0^{\pi} - \displaystyle\int_{0}^{\pi}2x\sin x\,dx$
Let $K=\displaystyle\int_{0}^{\pi}x\sin x\,dx$ By parts: $u=x,\; dv=\sin x\,dx \Rightarrow du=dx,\; v=-\cos x$
$K=[-x\cos x]_0^{\pi}+\displaystyle\int_{0}^{\pi}\cos x\,dx = \pi$
So $J=0-2K=-2\pi$
Then $\displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx = 3J = -6\pi$
$I = [-x^{3}\cos x]_0^{\pi} + 3J = (-\pi^{3}\cos\pi - 0) - 6\pi = \pi^{3} - 6\pi$
Answer: $\boxed{\pi^{3} - 6\pi}$ ✅
Let $t=1-x \Rightarrow x=1-t,\ dx=-dt$.
As $x:0\to1$,
$t:1\to0$.
\[ \int_{0}^{1} x(1-x)^n\,dx \]
\[ = \int_{1}^{0} (1-t)\,t^n\,(-dt) \]
\[= \int_{0}^{1} \big(t^n - t^{n+1}\big)\,dt \]
\[ = \left[\frac{t^{n+1}}{n+1}-\frac{t^{n+2}}{n+2}\right]_{0}^{1} = \frac{1}{n+1}-\frac{1}{n+2}. \]
Simplify: \[ \frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}. \]
Answer: $\boxed{\dfrac{1}{(n+1)(n+2)}}$ (valid for $n>-1$).
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