# I Odd/even function in integral

1. Sep 22, 2016

### KFC

I am reviewing some basic calculus with basic trigonometric functions. I remember for periodic function, one can use the feature of odd/even function to help computing some integral. I got two integrals from a book some times ago (I can't recall which book are they from). I expect those integrals to give zero from my note.

$\int_{-\pi}^\pi \frac{\cos[(m+1)(x+\pi/2)]\sin[m(x+\pi/2)]\sin(x)}{\cos(x)}dx$

Here $m$ is positive integer. On this integral, I think $\sin[m(x+\pi/2)]$ is even function because it shifts by $\pi/2$, the $\cos(x)$ and $\cos[(m+1)(x+\pi/2)]$ are also even functions except that $\sin(x)$ is an odd function, so the integrand is an odd function, the integral gives zero. I cannot find the result from the integral table and mathematica won't give me any number. So is my logic correct to get the zero?

$\int_{-\pi}^\pi \frac{\cos[(m+1)(x+\pi/2)]\sin[m(x+\pi/2)]}{\cos(x)}dx$

However, the similar reasoning fails in this case since all the even function. My note written that this integral should be zero but I cannot tell why.

I verify that those two integrals gives zero when m=1, 2, 3, ... 30 individually and manually. I didn't check the number after 30 since I am looking for a way to provide the general case for any m.

2. Sep 22, 2016

### andrewkirk

$\sin[m(x+\pi/2)]$ and $\cos[(m+1)(x+\pi/2)]$ are both only even if $m$ is odd, otherwise the functions are even.

3. Sep 22, 2016

### KFC

Thanks. Do you mean $\sin[m(x+\pi/2)] \cdot \cos[(m+1)(x+\pi/2)]$ will be odd function if $m$ is odd, the cosine part will be even and the sine part will be even too so their multiplication is odd; when $m$ is even, the cosine part is odd but the sine part is still odd, so their multiplication is always even.So it is possible to have even or odd function for the integrand. Even though we get the even one, it is not saying that the integration will be nonzero, we need to do more analysis for different m.

Last edited: Sep 22, 2016
4. Sep 22, 2016

### KFC

I am still looking for a way to prove hat those two integral gives zero. Following andrewkirk's comment, I am trying to break my proof into 4 integral by taking care of 4 different $m$ such that the following identities can be used

I1: $m=4t+1, \sin[m(x+\pi/2)] = \sin(mx+\pi/2) = \cos(mx); \cos[(m+1)(x+\pi/2)] = \cos[(m+1)x + \pi] = -\cos[(m+1)x]$
I2: $m=4t+2, \sin[m(x+\pi/2)] = \sin(mx+\pi) = -\sin(mx); \cos[(m+1)(x+\pi/2)] = \cos[(m+1)x + 3\pi/2] = \sin[(m+1)x]$
I3: $m=4t+3, \sin[m(x+\pi/2)] = \sin(mx+3\pi/2) = -\cos(mx); \cos[(m+1)(x+\pi/2)] = \cos[(m+1)x]$
I4: $m=4t+4, \sin[m(x+\pi/2)] = \sin(mx+2\pi) = \sin(mx); \cos[(m+1)(x+\pi/2)] = \cos[(m+1)x + \pi/2] = -\sin[(m+1)x]$

The integral about I1, I2 and I3 are proved to be zero by using the trigonometric identities and odd/even features. But I have a hard time to get the very last one proved. Here is what I did.

$\begin{eqnarray*} \int_{-\pi}^\pi \frac{\sin[m(x+\pi/2)]\cos[(m+1)(x+\pi/2)]}{\cos x}dx &=& -\int_{-\pi}^\pi \frac{\sin(mx)\sin[(m+1)x]}{\cos x}dx \\ &=& -\int_{-\pi}^\pi \frac{\sin(mx)[\sin(mx)\cos x + \cos(mx)\sin x]}{\cos x}dx\\ &=& -\int_{-\pi}^\pi \sin^2(mx)dx -\int_{-\pi}^\pi \frac{\sin(mx)\cos(mx)\sin x}{\cos x}dx \end{eqnarray*}$

In the first part of the very last expression, the integral is $-\pi$. To get the integral of the second part, I apply

$\sin(mx) = 2\cos x \sin[(m-1)x] - \sin[(m-2)x]$

$\begin{eqnarray*} \int_{-\pi}^\pi \frac{\sin(mx)\cos(mx)\sin x}{\cos x}dx &=& \int_{-\pi}^\pi \frac{\{2\cos x \sin[(m-1)x] - \sin[(m-2)x]\}\cos(mx)\sin x}{\cos x}dx \\ &=& 2{\int_{-\pi}^\pi \sin[(m-1)x]\sin x dx } - \int_{-\pi}^\pi \frac{\sin[(m-2)x]\cos(mx)\sin x}{\cos x}dx \end{eqnarray*}$

The first term is zero; apply the

$\cos(mx) = 2\cos x \cos[(m-1)x] - \cos[(m-2)x]$

to the second term, we can prove with similar trick that the integral is also zero. So it comes to my question, if that integral is zero, and $\int sin^2(mx)dx = \pi$, so I4 is not zero. But the book said it should be zero. What's wrong with my computation?

Last edited: Sep 22, 2016