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In my book there is an example where the definite integral of

$$\int\limits_0^{2\pi} \sin^2(nt) dt = \pi-\left[\frac{\sin(2n2\pi)}{4n} \right] $$

Then it says : $π$-$[sin(2*n*2π)/4*n ] $ = π

Why? If n is even,then yes, $[sin(2*n*2π)/4*n ] $ = 0 so $π$-$[sin(2*n*2π)/4*n ] $ = π

but in my book it does NOT say that n is even.So why do we take $π$-$[sin(2*n*2π)/4*n ] $ = π ?

DonAntonio
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fdxsd
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  • $\sin x = 0 \iff x = k\cdot \pi$ for some $k\in\mathbb{Z}$. $(2\cdot n\cdot 2\pi)$ is a multiple of $\pi$ for every $n\in \mathbb{Z}$. – Daniel Fischer Dec 16 '13 at 12:07
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    @fdxsd , please look at my editing of part of your question, check it is correct, try to learn the symbols and drop those annoying * to denote multiplication. – DonAntonio Dec 16 '13 at 13:22

1 Answers1

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Because for any $n$, even or odd, your argument $(2n\cdot 2\pi)$ is always an even number, and more importantly, it is always a multiple of $2\pi$. The sine of any multiple of $2\pi$ is 0, so $\pi-0=\pi$.

Eleven-Eleven
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