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Compute $\displaystyle \int \limits_0^\pi \sin(\sin(x))\sin(x)\mathrm dx$.

I have no idea how to integrate of this. I do need some help. Thanks

Git Gud
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Ison
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    As a starting point, you should surely try testing what happens if you take the derivatives of the four functions $\sin\sin x$, $\sin\cos x$, $\cos\sin x$, and $\cos\cos x$ just to see what happens. That would be the very first think I thought of. Then maybe you can see something you might be able to do to "undo" the derivative, or perhaps make a clever substitution. – MPW Mar 31 '14 at 21:51
  • WolframAlpha doesn't have a solution for it (for the indefinite integral). Therefore, it probably doesn't exist. However, the answer is about $1.3824597$. – Shahar Mar 31 '14 at 22:04
  • Try $\sin x = t$, and see what happends. – Imanol Pérez Arribas Mar 31 '14 at 22:06
  • It's apparently equal to $\pi J_1(1)$ which is some "Bessel function." Google told me that it was defined by Bernoulli. I quit, I can't be Bernoulli anymore. – Shahar Mar 31 '14 at 22:07
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    The function is symmetric about $\pi/2$, so the integral is the same as $2\int_0^{\pi/2}\sin(\sin(x))\sin(x),dx$. From there, a $u$-substitution and integration by parts show it is equal to $2\int_0^{\pi/2}\cos(u)\sqrt{1-u^2},du$, if that helps. – 2'5 9'2 Mar 31 '14 at 22:07
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    @ImanolPérezArribas As with Ayesha's answer, since that substitution is not invertible on $[0,\pi]$, that is not as helpful as it may seem. – 2'5 9'2 Mar 31 '14 at 22:08
  • @alex.jordan Well, I mean after dividing the integral into two different integrals: $\displaystyle \int \limits_0^\pi/2 \sin(\sin(x))\sin(x)\mathrm dx + \displaystyle \int \limits_{\pi/2}^\pi \sin(\sin(x))\sin(x)\mathrm dx$ – Imanol Pérez Arribas Mar 31 '14 at 22:11
  • @ImanolPérezArribas so that just leads to $2\int_0^{\pi/2}\sin(\sin(x))\sin(x), dx$: the same integrand over half the interval and then doubled. – 2'5 9'2 Mar 31 '14 at 22:12

3 Answers3

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Here's a solution using Taylor Series:

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

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

$$\int_0^\pi \sin(\sin x) \sin x \, dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\int_0^\pi(\sin x)^{2n+2} dx$$

Note that $\int_0^\pi(\sin x)^{2n+2} dx = \pi \frac{(2n+1)(2n-1)(2n-3)\cdots 3 \cdot 1}{(2n+2)(2n)(2n-2)\cdots 4 \cdot 2} = \pi \frac{(2n+1)!!}{(2n+2)!!}$ for any non-negative integer $n$. You can use the following reduction formula to prove it:

$$\int \sin^n x \, dx = - \frac{1}{n} \sin^{n-1} x \cos^{n-1} x + \frac{n-1}{n} \int \sin^{n-2} x \, dx$$

So we have:

$$\begin{align} \int_0^\pi \sin(\sin x) \sin x \, dx &= \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)} \frac{(2n+1)!!}{(2n+2)!!} \\ &= \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!!(2n+2)!!} \\ &= \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{(2^n n!) (2^{n+1} (n+1)!)} \\ &= \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+1} n! (n+1)!} \\ \end{align}$$

Now you can note that the summation here is exactly the definition of $J_1(1)$, where $J_\alpha(x)$ is the Bessel function of the first kind:

http://en.wikipedia.org/wiki/Bessel_function

So $$\int_0^\pi \sin(\sin x) \sin x \, dx = \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+1} n! (n+1)!} = \pi J_1(1) \approx 1.38246$$

Perry Elliott-Iverson
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  • +1 : in fact, IIRC all the Fourier coefficients of $\sin(\sin(x))$ can be expressed via Bessel functions - I actually came across this at one point in some audio research(!) into feedback circuits (specifically, finding the tones that appear when the output of a (constant-frequency) sine-wave generator is patched into the frequency input of another sine-wave generator). – Steven Stadnicki Mar 31 '14 at 23:05
  • You guys are awsome. Thanks a lot, but how about if the equation is changed to ∫sin(Asin(x))sin(x)dx, A is constant? – Ison Mar 31 '14 at 23:21
  • Why sin(sinx) and sinx have the same representation of taylor series? – Ison Mar 31 '14 at 23:45
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    For ∫sin(Asin(x))sin(x)dx, you will end up with another factor of $A^{2n+1}$ in the summation, which you can see would give you $\pi J_1(A)$ rather than $\pi J_1(1)$. As for your second comment, $\sin x$ and $\sin (\sin x)$ do not have the same representation as Taylor series. Note that I substituted $\sin x$ in for $x$ in the Taylor series representation of $\sin x$ to get a series representing $\sin (\sin x)$ (this technically isn't a Taylor series, since it has powers of $\sin x$ rather than powers of $x$. – Perry Elliott-Iverson Apr 01 '14 at 00:05
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Hint: What happens when we try the substitution $u = \sin{x}$?

Ayesha
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    What does happen? – Git Gud Mar 31 '14 at 21:53
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    You end up with $2\int_{0}^{1}\frac{u\sin u}{\sqrt{1-u^2}}du = \pi\cdot J_1(1),$ so the integral can be written in terms of a specific value of a Bessel function. – Jack D'Aurizio Mar 31 '14 at 22:11
  • By writing $\sin(\sin x)$ as $\sum_{j=0}^{+\infty}\frac{(-1)^j (\sin x)^{2j+1}}{(2j+1)!}$ and integrating termwise, we get the pretty fast convergent series: $$\int_{0}^{\pi}\sin(\sin x)\sin x,dx = \pi\cdot J_1(1) = \pi\sum_{j=0}^{+\infty}\frac{(-1)^j}{j!(j+1)!2^{2j+1}}.$$ – Jack D'Aurizio Mar 31 '14 at 22:19
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This reminds me of

$ \frac{1}{\pi}\int_{0}^{\pi} \sin(\cos(\sin(\cos( x)))) dx $ where we iterate the expression in the integrand, replacing $x$ by $ \sin(\cos( x)) $ and which goes to $0.6948196907307875\ldots$ (a fixed point of $\sin(\cos(x_0)) = x_0$).

max_zorn
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Alan
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