En lugar de ello, vamos a utilizar la inducción y (iterada) integración por partes. (Tenga en cuenta que no debe ser un "${}+C$ " en esa fórmula, pero no me voy a preocupar de eso.)
En primer lugar, vamos a tomar ninguna $n\ge 1$ e integrar a $\int x^n\sin x\,dx$ por partes a ver qué pasa. Por el LIATE Regla, debemos tomar las $u_1=x^n$$dv_1=\sin x\,dx$, dándonos $du_1=nx^{n-1}\,dx$$v_1=-\cos x$. Entonces $$\int x^n\sin x\,dx=\int u_1\,dv_1=u_1v_1-\int v_1\,du_1=-x^n\cos x+n\int x^{n-1}\cos x\,dx.$$ The integral on the far right is easy when $n=1$, but if $n\ge 2$ then it's only slightly less problematic than the integral we started with. Still, it's an improvement, so we'll bear it in mind: $$\int x^n\sin x\,dx=-x^n\cos x+n\int x^{n-1}\cos x\,dx\quad\quad\text{for }n\ge 1.\tag{1}$$ Now let's suppose $n\geq 2$, and integrate $\int x^{n-1}\cos x\, dx$ by parts. Take $u_2=x^{n-1}$ and $dv_2=\cos x\,dx$, so $du_2=(n-1)x^{n-2}\, dx$ and $v_2=\sin x$. Then $$\int x^{n-1}\cos x\,dx=\int u_2\,dv_2=u_2v_2-\int v_2\,du_2=x^{n-1}\sin x-(n-1)\int x^{n-2}\sin x\,dx,$$ so by $(1)$, we have $$\begin{align}\int x^n\sin x\, dx &= -x^n\cos x+n\int x^{n-1}\cos x\,dx\\ &= -x^n\cos x+nx^{n-1}\sin x-n(n-1)\int x^{n-2}\sin x\,dx.\end{align}$$ Hence, we've rewritten the original integral in terms of polynomial combinations of $\el pecado x$ and $\cos x$, together with an integer multiple of an integral much like the one we started with, but with a power of $x$ that is $2$ smaller. This will allow us to make an inductive argument, but with jumps of $2$, so we'll need $2$ base cases instead of $1$. Namely, we'll need base cases $n=1,2$, and we'll induce along the odd $n$ and the even $n$ separately. Let's bear it in mind: $$\int x^n\sin x\, dx=-x^n\cos x+nx^{n-1}\sin x-n(n-1)\int x^{n-2}\sin x\,dx\quad\text{for }n\ge 2.\tag{2}$$
Para el $n=1$ de los casos, simplemente podemos utilizar $(1)$ para obtener $$\int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x.$$ On the other hand, the following $4$ las líneas son todos iguales:
$$\sum_{k=0}^{\lfloor{1/2}\rfloor}(-1)^{k+1}x^{1-2k}{1!\over(1-2k)!}\cos x+\sum_{k=0}^{\lfloor{(1-1)/2}\rfloor}(-1)^kx^{1-2k-1}{1!\over(1-2k-1)!}\sin x$$
$$\sum_{k=0}^0(-1)^{k+1}x^{1-2k}{1\over(1-2k)!}\cos x+\sum_{k=0}^0(-1)^kx^{-2k}{1\over(-2k)!}\sin x$$
$$(-1)^{0+1}x^{1-0}{1\over(1-0)!}\cos x+(-1)^0x^{0}{1\over(0)!}\sin x$$ $$-x\cos x+\sin x,$$ así que estamos bien en el primer caso base.
Para el $n=2$ de los casos, se puede igualmente utilizar $(2)$ y calcular las sumas de forma explícita para confirmar que la fórmula se mantiene.
Ahora, vamos a hacer el impar de inducción. Estamos considerando a todos $n=2m-1$ ($m\in\Bbb N$). Vamos a sustituir en la fórmula deseada para conseguir lo que estamos tratando de demostrar que en términos de $m$, en su lugar. La observación de que ese $\lfloor\frac{n}2\rfloor=\lfloor m-\frac12\rfloor=m-1$ $\lfloor\frac{n-1}2\rfloor=\lfloor m-1\rfloor=m-1,$ queremos mostrar que $$\begin{align}\int x^{2m-1}\sin x\,dx=\sum_{k=0}^{m-1}(-1)^{k+1}x^{2m-1-2k}{(2m-1)!\over(2m-1-2k)!}\cos x\\+\sum_{k=0}^{m-1}(-1)^kx^{2m-2-2k}{(2m-1)!\over(2m-2-2k)!}\sin x\end{align}\tag{3}$$ for all $m\in\Bbb, N$. We already know the formula holds in the $m=1$ ($n=1$) case, and by $(2)$, we have $$\begin{align}\int x^{2(m+1)-1}\sin x\, dx=-x^{2(m+1)-1}\cos x+\bigl(2(m+1)-1\bigr)x^{2(m+1)-2}\sin x\\-\bigl(2(m+1)-1\bigr)\bigl(2(m+1)-2\bigr)\int x^{2m-1}\sin x\,dx\end{align}\tag{4}$$ for all $m\geq 1$.
Supongamos que para algunos $m$ que $(3)$ mantiene. Tenga en cuenta que para cualquier función de $f(x)$,$\sum\limits_{k=0}^{m-1}f(k)=\sum\limits_{k=1}^mf(k-1)$. El uso de esta nueva indexación de truco, tenemos por $(3)$ que $$\begin{align}\int x^{2m-1}\sin x\,dx &=\sum_{k=0}^{m-1}(-1)^{k+1}x^{2m-1-2k}{(2m-1)!\over(2m-1-2k)!}\cos x\\ &{}\qquad+\sum_{k=0}^{m-1}(-1)^kx^{2m-2-2k}{(2m-1)!\over(2m-2-2k)!}\sin x\\ &=\sum_{k=1}^{m}(-1)^{(k-1)+1}x^{2m-1-2(k-1)}{(2m-1)!\over(2m-1-2(k-1))!}\cos x\\ &{}\qquad+\sum_{k=1}^{m}(-1)^{k-1}x^{2m-2-2(k-1)}{(2m-1)!\over(2m-2-2(k-1))!}\sin x\\ &=(-1)^{-1}\sum_{k=1}^{m}(-1)^{k+1}x^{2m+1-2k}{(2m-1)!\over(2m+1-2k))!}\cos x\\ &{}\qquad+(-1)^{-1}\sum_{k=1}^{m}(-1)^{k}x^{2m-2k}{(2m-1)!\over(2m-2k)!}\sin x\\ &=-\sum_{k=1}^{(m+1)-1}(-1)^{k+1}x^{2(m+1)-1-2k}{(2(m+1)-3)!\over(2(m+1)-1-2k))!}\cos x\\ &{}\qquad-\sum_{k=1}^{(m+1)-1}(-1)^{k}x^{2(m+1)-2k}{(2(m+1)-3)!\over(2(m+1)-2-2k)!}\sin x,\end{align}$$ and so $$\begin{align}-\bigl(2(m+1)-1\bigr)\bigl(2(m+1)-2\bigr)\int x^{2m-1}\sin x\,dx =\sum_{k=1}^{(m+1)-1}(-1)^{k+1}x^{2(m+1)-1-2k}{(2(m+1)-1)!\over(2(m+1)-1-2k))!}\cos x\\ {}\qquad+\sum_{k=1}^{(m+1)-1}(-1)^{k}x^{2(m+1)-2k}{(2(m+1)-1)!\over(2(m+1)-2-2k)!}\sin x.\end{align}$$ Thus, since $$-x^{2(m+1)-1}\cos x=(-1)^{0+1}x^{2(m+1)-1-0}{(2(m+1)-1)!\over(2(m+1)-1-0)!}\cos x$$ and $$(2(m+1)-1)x^{2(m+1)-2}\sin x=(-1)^{0}x^{2(m+1)-2-0}{(2(m+1)-1)!\over(2(m+1)-2-0)!}\sin x,$$ we have by $(4)$ and the above work that $$\begin{align}\int x^{2(m+1)-1}\sin x\,dx=\sum_{k=0}^{(m+1)-1}(-1)^{k+1}x^{2(m+1)-1-2k}{(2(m+1)-1)!\over(2(m+1)-1-2k)!}\cos x\\+\sum_{k=0}^{(m+1)-1}(-1)^kx^{2(m+1)-2-2k}{(2(m+1)-1)!\over(2(m+1)-2-2k)!}\sin x,\end{align}$$ and so the desired formula holds in the $m+1$ caso, también.
La inducción, vamos a proceder de una manera similar a la extraña inducción, excepto que vamos a estar considerando la posibilidad de $n=2m$$m\in\Bbb N$.
