# Integrals involving irrational numbers

$${1 \over 2\pi} \int_{0}^{2\pi} \sin^{100}(x) dx$$

How should I approach getting an estimate on the numeric value in this case?

• What are allowed to use? A calculator? A calculator with a numeric integrator built-in? Jan 26 at 12:56
• Closely related: math.stackexchange.com/questions/24533/… Jan 26 at 13:06

Due to the symmetry of the graph of $$\sin^{100} x$$, the integral equals $$\frac{4}{2\pi} \int_0^{\pi/2} \sin^{100} x \ dx =\frac{2}{\pi} \cdot \frac{100!}{(2^{50}50!)^2}\cdot \frac{\pi}{2}= \frac{100!}{2^{100} \cdot (50!)^2}$$

using this result.

Probably, you should use complex numbers. Say, $$\frac{1}{2 \pi} \int_{0}^{2 \pi} \sin^{100}(x) dx = \frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{(e^{ix} - e^{-ix})^{100}}{2^{100}} dx = \frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{e^{-100 i x}}{2^{100}}(e^{2 ix} - 1)^{100} dx.$$ If we exand $$(e^{2 ix} - 1)^{100}$$ and start doing integrals, we will see that they all are zero except $$\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{e^{-100 i x}}{2^{100}} {100\choose 50} e^{2 \times 50 \times ix} dx = \frac{1}{2^{100}} {100\choose 50} \approx 0.0795...$$

• Excellent answer Mar 4 at 13:30

Integrate by parts successively to reduce the integral

$$I_{100}= \int_{0}^{2\pi} \sin^{100} x dx =-\frac1{100}\int_{0}^{2\pi} \tan^{99} x \>d(\cos^{100}x)\\=\frac{99}{100}I_{98} =\frac{99}{100} \frac{97}{98} I_{96}= \cdots =\frac{99}{100} \frac{97}{98} \cdots \frac12\cdot2\pi$$

Use Mathematica [1]: $$\frac{1}{2\pi} \int_0^{2 \pi}\sin^{100}x\, dx = \frac{12611418068195524166851562157}{158456325028528675187087900672}\approx 0.0795892$$

• OP probably would not be allowed to use Mathematica nor Wolfram Alpha, or else they would have used it, nor does your answer provide any insight as to how your fraction was derived (unlike other answers). This doesn't seem particularly useful.
– Kyky
Jan 26 at 15:22
• @Kyky I’m am simply answering OP’s question. You are making assumptions on what OP may or may not do. I don’t understand the downvotes. Jan 26 at 18:30