Calcular la integral indefinida $\int x^n \sin x\,dx$

Question

$\newcommand{\term}[3]{ \sum_{k=0}^{\lfloor #1/2 \rfloor} (-1)^{#2} x^{#3} \frac{n!}{(#3)!} }$ Estoy tratando de demostrar que para $n \in\mathbb N$, $$\int x^n \sin x \, dx = \cos x \término{n}{k+1}{n-2k} + \sin x \término{(n-1)}{k}{n-2k-1}$$ Empecé con la diferenciación, y esto es lo que obtuve:

Si definimos $f(x)$ $$f(x) = \cos x \término{n}{k+1}{n-2k} + \sin x \término{n-1}{k}{n-2k-1}$$ entonces tenemos $$\begin{align*} f'(x) &= \cos x \term{n}{k+1}{n-2k-1} - \sin x \term{n}{k+1}{n-2k} \\ &\qquad + \sin x \term{(n-1)}{k}{n-2k-2} + \cos x \term{(n-1)}{k}{n-2k-1} \\[8pt] &= \cos x \left[ \term{n}{k+1}{n-2k-1} + \term{(n-1)}{k}{n-2k-1} \right] \\ &\qquad + \sin x \left[ \term{(n-1)}{k}{n-2k-2} - \term{n}{k+1}{n-2k} \right] \end{align*}$$ No sé cómo seguir, porque de los diferentes límites de la suma con $\lfloor{n/2}\rfloor$$\lfloor{(n-1)/2}\rfloor$.

Answer 1

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.)

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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: $$\int x^n\sin x\, dx=-x^n\cos x+nx^{n-1}\sin x-n(n-1)\int x^{n-2}\sin x\,dx\quad\text{for }n\ge 2.\tag{2}$$

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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.

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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.

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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.

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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$.

Answer 2

Por fin tengo mi prueba con la diferenciación terminado, demasiado.

Si $$f(x) = \sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k}{n!\over(n-2k)!}\cos x+\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^kx^{n-2k-1}{n!\over(n-2k-1)!}\sin x$$

con $n\in \Bbb N$.

entonces $$\begin{align}f'(x) &= \left(\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}\right)\cos x\\&{}\quad+\left(\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k}x^{n-2k-2}{n!\over(n-2k-2)!}-\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k}{n!\over(n-2k)!}\right)\sin x\end{align}$$

$$\begin{align}&= \left(\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}\right)\cos x\\&{}\quad+\left(\sum_{k=2}^{\lfloor{(n-1)/2}\rfloor+2}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}-\sum_{k=1}^{\lfloor{n/2}\rfloor+1}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}\right)\sin x\end{align}$$

$$\begin{align}&=\left(\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}\right)\cos x\\ &{}\quad+\left(\sum_{k=2}^{\lfloor{(n-1)/2}\rfloor+2}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}-\sum_{k=2}^{\lfloor{n/2}\rfloor+1}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}\right)\sin x\\ &{}\quad+ x^n\sin x\end{align}$$

Ahora tenemos que demostrar que (1) $$\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!} = 0$$ and (2) $$\sum_{k=2}^{\lfloor{(n-1)/2}\rfloor+2}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}-\sum_{k=2}^{\lfloor{n/2}\rfloor+1}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!} = 0$$

Incluso los números: $$\lfloor{n/2}\rfloor = n/2$$ $$\lfloor{(n-1)/2}\rfloor = (n-2)/2$$

Para los números impares: $$\lfloor{n/2}\rfloor = (n-1)/2$$ $$\lfloor{(n-1)/2}\rfloor = (n-1)/2$$

También se $n!$ sólo está definida para $$n\ge 0$$

(1) El extraño caso es trivial debido a que de $$\lfloor{(n-1)/2}\rfloor = \lfloor{n/2}\rfloor$$

Incluso: necesitamos $$n -2k -1 \ge 0$$ lo $$k \le \lfloor{(n-1)/2}\rfloor = (n-2)/2$$

Obtenemos: $$\sum_{k=0}^{(n-2)/2}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{(n-2)/2}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!} = 0$$

Con la misma argumentación se puede mostrar (2).

Sigue: $$= 0\cos x+0\sin x+ x^n\sin x= x^n\sin x$$

Answer 3

Integrar por partes dos veces, y el uso de dos inducciones (pares e impares caso). Usted puede ser capaz de unir a estos si son hábiles.