I'm working on a problem for university Calculus 2. We're talking about Taylor series right now and I need to approximate an integral using one of a function that I think it should be easy to produce a series for, but I'm not 100% sure. This is the function:

$$f(x) = 2x\sin(x)$$

I know the expansion for $\sin(x)$, which is in a reference table in the book. To give the first few terms it looks like this:

$$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \;\cdots$$

I'm pretty sure I can just multiply the whole polynomial by $2x$, giving:

$$2x^2 - \frac{2x^4}{3!} + \frac{2x^6}{5!} - \frac{2x^8}{7!} + \;\cdots$$

What's odd is that I can't find any examples quite like this in either of the textbooks or on the internet... which makes me wonder if this isn't actually a valid manipulation. Additionally, I can't seem to get an answer that matches this from Wolfram Alpha.

I could work out the Taylor series by hand, but the derivatives of the function start getting a bit ugly (by which I mean long), so I think I'm supposed to manipulate a known series since this should be an easy problem.

So, am I doing this right or am I on the wrong track?


Your approach is correct. WolframAlpha returns the same series for the following query: Series[2 x Sin[x],{x,0,8}]


In order to estimate the integral of $2x\sin x$, you use the series for $\sin$ as you wrote above, multiplied by $2$ and $x$ as you did.


$$2x\sin x =2\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)!}= 2x^2 - \frac{2x^4}{3!} + \frac{2x^6}{5!} - \frac{2x^8}{7!} \cdots $$

We then integrate the summand or term-by-term to get

$$ 2\int x\sin x\, dx= 2\sum_{n=0}^{\infty} \int\frac{(-1)^n x^{2n+2}}{(2n+1)!}\,dx= 2\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)!}\int x^{2n+2}\,dx= 2\sum_{n=0}^{\infty}\frac{(-1)^n x^{3+2 n}}{(3+2 n) (1+2 n)!} $$

This sum can be used to estimate the integral of $2x\sin x$ to a desired accuracy.

To check your result,

$$2\int x\sin x=-2 (x \cos(x)-\sin(x))$$

  • $\begingroup$ It is important to note that you can integrate termwise because the series is absolutely convergent in $\mathbb R$ $\endgroup$ – Pedro Tamaroff Apr 29 '12 at 22:03

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