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