Odd function in an integral I have the following integral
\begin{equation}
\int^{2\pi}_{0}(16 \sin \phi \cos^2\phi-4\sin^{2}\phi+8\sin\phi)d\phi = -4\pi
\end{equation}
It says that is odd function of $\phi$. Can anyone explain?
 A: The expression $(16 \sin \phi \cos^2\phi-4\sin^{2}\phi+8\sin\phi)$ is not an odd function of $\phi$.
Consider each of the terms one-by-one:


*

*\begin{equation}
\int^{2\pi}_{0} 16 \sin \phi \cos \phi\ d \phi = 0
\end{equation}
because $\sin \phi \cos \phi$ is antisymmetric (odd) about the $\phi=\pi$ line.

*\begin{equation}
\int^{2\pi}_{0} -4\sin^{2}\phi\ d\phi\ \neq 0
\end{equation}
because $\sin^{2}\phi$ is symmetric (even) about the $\phi=\pi$ line.

*\begin{equation}
\int^{2\pi}_{0} 8\sin\phi\ d\phi= 0
\end{equation}
because $\sin\phi$ is antisymmetric (odd) about the $\phi=\pi$ line.
Therefore, we only need to calculate the following:
\begin{equation} \int^{2\pi}_{0} -4\sin^{2}\phi\ d\phi\  = 2 \int^{\pi}_{0} -4\sin^{2}\phi\ d\phi\ = - 4 \pi \end{equation}.
A: The above answer is very good to understand why the integrand is odd. I will instead focus on how to prove the value of the integral. Note that for all integers $m,n\in\mathbb{Z}$ one have
$$
    \int_0^{2\pi} \sin mx \cos nx \,\mathrm{d}x = 0
$$
This can be shown by rewriting 
$$
    \sin mx \cos nx = \frac{1}{2}\sin(m+n) x - \frac{1}{2}\sin(m-n) x
$$
However one can also look at this as $\sin x$ and $\cos x$ being orthogonal. Now similarly since $\sin x$ and $\cos x$ are symmetric about $x = \pi$ we have $\int_0^{2\pi} \sin x\,\mathrm{d}x = \int_0^{2\pi} \cos x\,\mathrm{d}x = 0$. Try to study their graphs. 
A bit harder, but by studying the graphs of $\cos^2x$ and $\sin^2x$ one can come to the conclusion that
$$
  \int_0^{2\pi} (\cos x)^2\,\mathrm{d}x
= \int_0^{2\pi} (\sin x)^2\,\mathrm{d}x
$$
I can show this more in detail, but do not want to detrail from the progress in the answer.
Now by using $(\cos x)^2 + (\sin x)^2 = 1$ we have
$$
\begin{align*}
2\int_0^{2\pi} (\sin x)^2\,\mathrm{d}x
& = \int_0^{2\pi} (\sin x)^2\,\mathrm{d}x + \int_0^{2\pi} (\sin x)^2\,\mathrm{d}x \\
& = \int_0^{2\pi} (\sin x)^2\,\mathrm{d}x + \int_0^{2\pi} (\cos x)^2\,\mathrm{d}x \\
& = \int_0^{2\pi} 1 \,\mathrm{d}x 
  = 2\pi 
\end{align*}
$$
So by conclusion we can say that Integrating $(\cos x)^2$ or $(\sin x)^2$ over an interval which is a multiple of $\pi/2$ is the same as integrating $1/2$ over the same interval. In conclusion we have
$$
\begin{align*}
I & = \int^{2\pi}_{0}16 \sin \phi \cos\phi-4\sin^{2}\phi+8\sin\phi\,\mathrm{d}\phi \\
& = 16 \cdot 0 - 4 \int^{2\pi}_{0} \frac{1}{2} \,\mathrm{d}\phi + 8 \cdot 0\\
& = - 4\pi
\end{align*}
$$
Which is what we wanted to show.
