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I would like to evaluate the following integral using differentiation under the integral sign.

$$\int\limits_{0}^{2\pi}\sin^{8}(x)dx$$

Unfortunately, I can't come up with a proper choice for a function with a parameter. $\sin^{8}(ax)$ won't work out, neither will $\sin^{a}(x)$... So maybe someone could give me a hint to make an appropriate choice of that function.

Thanks!

A-Level Student
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GuntherOnFire
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    I see no reason to expect the Leibniz rule will help evaluate this integral. https://en.wikipedia.org/wiki/Leibniz_integral_rule – GEdgar May 20 '21 at 11:03
  • Could you explain why please? I don’t see any violations. The integrand is continuous, the partial derivative is continuous, too, everything is well-behaved. So if I find an appropriate continuous parametrized function to get the desirable pattern, it’s gonna be fine, isn’t it? – GuntherOnFire May 20 '21 at 11:28
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    You will need a parameterization where the derivative with respect to the parameter is easy to integrate from $0$ to $2\pi$. The things i tried all produced more complicated integrands when differentiated. – GEdgar May 20 '21 at 11:31
  • Oh, I see... I thought maybe there are some restrictions I’ve missed. That 8th power is a problem, yes. Whatever, I’m just curious whether it is possible to somehow solve this one by means of this technique. – GuntherOnFire May 20 '21 at 11:41
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    It's not just about the conditions when differentiation under the integral sign is valid. Think about when this trick is helpful - it reduces an order of a power. In this case you would need a function such that $f^{(8)}(a \sin x)$, $f^{(4)}(a\sin^2x)$, $f^{(2)}(a\sin^4x)$, or $f'(a\sin^8x)$ are your integrand. Do you see why the technique would be unhelpful? – Ninad Munshi May 20 '21 at 11:41
  • Well, I’m not sure... I would say that whatever is a function I choose (at least of those that I tried), there will always be that 8th power and I’ll have to integrate it. If it’s not what you mean, then I’d appreciate if you explained it. – GuntherOnFire May 20 '21 at 12:02
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    You can use the reduction formula which will reduce the power by $2$ each time you apply it. – Toby Mak May 20 '21 at 12:25

2 Answers2

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Call the integral $I$. Note that

$$I = \int_0^{2\pi} \sin^8{x} \, \mathrm{dx} = 4\int_0^{\pi/2} \sin^8{x} \, \mathrm{dx}$$

Let $x = \arctan{t}$. Then

$$I = 4\int_0^\infty \frac{t^8}{(1+t^2)^5} \, \mathrm dt.$$

Define $$f(\alpha) = 4\int_0^\infty \frac{1}{(1+\alpha t^2)} \, \mathrm dt = \frac{2\pi}{\sqrt{\alpha}}$$

Taking the fourth derivative of both sides we have:

$$ f^{(4)}(\alpha) =24 \int_0^\infty \frac{4t^8}{(1+\alpha t^2)^5} \, \mathrm dt = \frac{105\pi}{8\sqrt{\alpha^9}}$$

$$ I = \frac{1}{24} f^{(4)}(1) = \frac{1}{24} \cdot \frac{105\pi}{8} = \frac{35\pi}{64}.$$

The easiest way to to see that $$\displaystyle \displaystyle I = 4\int_0^{\pi/2} \sin^8 x \, \mathrm dx.$$

is to look at the graph of $f(x) = \sin^8 x$. The area under the curve from $0$ to $2\pi$ is 4 times the area under the curve from $0$ to $\pi/2$. Alternatively, we can derivative this algebraically by splitting the integral:

$\displaystyle I = \int_0^{\pi/2} \sin^8 x \, \mathrm dx + \int_{\pi/2}^{\pi} \sin^8 x \, \mathrm dx+\int_\pi^{3\pi/2}\sin^8 x \, \mathrm dx+\int_{3\pi/2}^{2\pi/2} \sin^8 x \, \mathrm dx$

Let $t = x-\pi/2$, $t = x-\pi$, $t = x-3\pi/2$ for the $2$nd, $3$rd and $4$th integrals:

$\displaystyle I = \int_0^{\pi/2} \sin^8 t \, \mathrm dt + \int_{0}^{\pi/2} \cos^8 t \, \mathrm dt+\int_{0}^{\pi/2}\sin^8 t \, \mathrm dt+\int_{0}^{\pi/2} \cos^8 t \, \mathrm dt$

Let $t = \pi/2-u$ for the $2$nd/$4$th integrals then you get

$$\displaystyle \displaystyle I = 4\int_0^{\pi/2} \sin^8 t \, \mathrm dt.$$

J.G.
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Zack
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    Thank you so much! That’s really amazing! The only thing that’s not quite clear is moving from one integral from 0 to $2\pi$ to four integrals from 0 to $\frac{\pi}{2}$. Could you explain this step please? I understand it’s an even function, yet I can’t see why we can make it that way. – GuntherOnFire May 20 '21 at 19:02
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    You can observe that from the graph of the function $f(x) = \sin^ 8 x $. It's a little bit longer but you can also derive it algebraically. See the added explanation. – Zack May 21 '21 at 02:51
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    Thanks so much! Everything’s perfectly clear now! – GuntherOnFire May 21 '21 at 03:40
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    How did you come up with this brilliant answer? Random experimentation? – A-Level Student May 30 '21 at 10:57
  • @A-LevelStudent Thank you. I had to change the integral into another form since it wasn't clear how to insert a useful parameter into it. After that it was just experience because I've used the 'differentiate both sides' trick before on integrals of rational functions. – Zack May 31 '21 at 02:14
  • @NoName I see, thanks for the explanation, I guess I'll practice it a lot more and try to get as good at it as you are :) – A-Level Student May 31 '21 at 20:53
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You could try to use the fact that: $$\cos(8x)=\Re\left(\left[\cos x+j\sin x\right]^8\right)$$ And rearrange. Then use the half angle formula, although I think some of the terms would still be quite ugly

Henry Lee
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