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I haven't formally learned integrals, but I was trying to apply what I do know.

Is $$\int_{a}^{b}m\sin(k(x+j)) dx =0$$

as long as $b-a\equiv 0 \pmod {\frac{2\pi}{k}-j}$

and $m, k, j \in \mathbb{R}$?

So I'm trying to see if I can generalize when the positive and negative curves of a sinusoid will cancel each other out and make the integral 0.

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The "$j$" part is not relevant. Certainly you will get $0$ if $b-a$ is $\frac{2\pi}{k}$ times an integer. There are other situations in which you get $0$, because of the shape of the cosine function. – André Nicolas Oct 29 '12 at 17:44
Why isn't the j relevant? If it was shifted left or right, wouldn't that change where the halves line up? – Maddy Byahoo Oct 29 '12 at 18:02
If you move a bit to the right, you "lose" some area at the left end, but "gain" the same amount at the right end. – André Nicolas Oct 29 '12 at 18:06
up vote 1 down vote accepted

We will use the calculus. A non-calculus argument can be given, but it would involve drawing a number of pictures, a painful process.

Suppose that $k\ne 0$. Then a definite integral of $\sin(k(x+j))$ is $-\frac{1}{k}\cos(k(x+j))$. It follows that our integral is $0$ precisely if $\cos (k(a+j))=\cos(k(b+j))$.

There are two sorts of situations where $\cos u=\cos v$. The simplest is when $u$ and $v$ differ by a multiple of $2\pi$. So our integral will be $0$ if $k(b+j)-k(a+j)=2n\pi$ for some integer $n$. This is the case if $k(b-a)=2n\pi$, or equivalently if $$b-a=n\frac{2\pi}{k}$$ for some integer $n$. The congruence notation is most often used for integers. However, it can be extended, and we can write $x\equiv y\pmod{t}$ if $x-y$ is an integer multiple of $t$. Then the above result can be expressed as $b\equiv a\pmod{\frac{2\pi}{k}}$.

Note that $\cos(w)=\cos(-w)$. So we can also have $\cos u=\cos v$ if the sum of $u$ and $v$ is an integer multiple of $2\pi$. In our case, that gives the condition that $k(a+b)+2kj$ is a multiple of $2\pi$, or, in your congruence notation, $a+b+2j\equiv 0\pmod{\frac{2\pi}{k}}$. Unlike in the previous case, here the value of $j$ is relevant.

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I'm not sure why the integral of $sin(k(x+j))$ is $\frac{-1}{k}cos(k(x+j))$. Where is the 1/k coming from? – Maddy Byahoo Oct 30 '12 at 16:28
If you have done integration, make the substitution $u=k(x+j)$. But more simply, if you know how to differentiate, find the derivative of $-\frac{1}{k}\cos(k(x+j))$, using the Chain Rule. You will find that the result is $\sin(k(x+j))$. – André Nicolas Oct 30 '12 at 16:34

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