Definite integral of the product of powers of the sine and cosine When integration limits were from $0$ to $2\pi$ and the function was $\sin x \cdot \cos^2 x$ and also when the function was $\cos x \cdot \sin^2 x$,answer is $0$, so is there a standard formula when limits are $0$ to $2\pi$ and function is product of higher powers of $\sin x$ and $\cos x$? I have searched about it but got no such information..
 A: Here's an answer not just for higher powers of $\sin x$ and $\cos x,$
but indeed for any non-negative integer powers of $\sin x$ and $\cos x.$
In this answer, assume at all times that $m$ and $n$ are non-negative integers.
If $m$ and $n$ both are even, then
$$
\int_0^{2\pi} \sin^m t \cos^n t\,dt
$$
evaluates to a positive real number according to procedures shown later in this answer.
In any other case ($m$ odd, $n$ odd, or $m$ and $n$ both odd),
$$
\int_0^{2\pi} \sin^m t \cos^n t\,dt = 0.
$$

Consider three cases that cover all possibilities:
Case 1: $m$ is odd.
Since $\sin(-t) = -\sin t$ and $m$ is odd, $\sin^m(-t) = -\sin^m t,$
that is, $\sin$ is an odd function.
But $\cos(-t) = \cos(t)$ and therefore $\cos^n(-t) = \cos^n t.$
It follows that
\begin{align}
\sin^m(-t)\cos^n(-t) &= -\sin^m t \cos^n t; \\
\int_{-\pi}^0 \sin^m t \cos^n t\,dt &= - \int_0^\pi \sin^m t \cos^n t\,dt; \\
\int_{-\pi}^\pi \sin^m t \cos^n t\,dt &= 0. \\
\end{align}
Now observe that
$\sin(t - 2\pi) = \sin t$
and $\cos(t - 2\pi) = \cos t,$
so $\sin^m (t - 2\pi) \cos^n (t - 2\pi) = \sin^m t \cos^n t$
and
\begin{align}
\int_0^{2\pi} \sin^m t \cos^n t\,dt
& = \int_0^\pi \sin^m t \cos^n t\,dt
     + \int_\pi^{2\pi} \sin^m t \cos^n t\,dt \\
& = \int_0^\pi \sin^m t \cos^n t\,dt
     + \int_{-\pi}^0 \sin^m t \cos^n t\,dt \\
&= \int_{-\pi}^{\pi} \sin^m t \cos^n t\,dt.
\end{align}
Therefore
$$ \int_0^{2\pi} \sin^m t \cos^n t\,dt = 0. $$
Case 2: $n$ is odd.
Since (as we already know) 
$\sin^m (t - 2\pi) \cos^n (t - 2\pi) = \sin^m t \cos^n t,$
it follows that
\begin{align}
\int_0^{2\pi} \sin^m t \cos^n t\,dt
& = \int_0^{3\pi/2} \sin^m t \cos^n t\,dt
     + \int_{3\pi/2}^{2\pi} \sin^m t \cos^n t\,dt \\
& = \int_0^{3\pi/2} \sin^m t \cos^n t\,dt
     + \int_{-\pi/2}^0 \sin^m t \cos^n t\,dt \\
&= \int_{-\pi/2}^{3\pi/2} \sin^m t \cos^n t\,dt \\
&= \int_0^{2\pi} \sin^m \left(t-\frac\pi2\right)
                 \cos^n \left(t-\frac\pi2\right)\,dt \\
&= \int_0^{2\pi} \left(-\cos t\right)^m \sin^n t\,dt \\
&= \pm \int_0^{2\pi} \cos^m t \sin^n t\,dt \\
\end{align}
(where the $\pm$ sign depends on whether $m$ is even or odd).
But we now have $\sin t$ raised to an odd power, so Case 1 shows that
$$ \int_0^{2\pi} \cos^m t \sin^n t\,dt = 0.$$
Therefore
$$ \int_0^{2\pi} \sin^m t \cos^n t\,dt = 0.$$
Case 3: $m$ and $n$ are both even.
In this case we can use the identity $\sin^2 t = 1 - \cos^2 t$ to show that
$$
\int_0^{2\pi} \sin^m t \cos^n t\,dt
 = \int_0^{2\pi} (1 - \cos^2 t)^{m/2} \cos^n t\,dt,
$$
and since $\frac m2$ is an integer we can expand $(1 - \cos^2 t)^{m/2}$
to get a polynomial in $\cos^2 t.$
Since $n$ is even, $\cos^n t$ is a power of $\cos^2 t$
and therefore the entire integrand
$$ (1 - \cos^2 t)^{m/2} \cos^n t $$
is a polynomial in $\cos^2 t.$
We can integrate each term of this polynomial separately;
the integral for any non-constant term can be evaluated using the formula
$$
\int_0^{2\pi} \cos^{2p} t\,dt 
= 2\pi \left(\frac{1 \cdot 3 \cdot 5 \cdots(2p-1)}{2 \cdot 4 \cdot 6 \cdots 2p}\right)
$$
where $p$ is a positive integer
(see Definite integral of even powers of Cosine.,
integration of $\int_0^{2\pi} cos^{2n}(t)dt$,
and the answers to those questions for details).
In the case $n=0$ there is also a constant term whose integral
(of course) is $\int_0^{2\pi} 1\,dt = 2\pi.$
Example: Integrate $\sin^4 t \cos^2 t.$
We have
\begin{align}
\int_0^{2\pi} \sin^4 t \cos^2 t \,dt 
&= \int_0^{2\pi} (1 - \cos^2 t)^2 \cos^2 t \,dt \\
&= \int_0^{2\pi} (\cos^2 t - 2\cos^4 t + \cos^6 t)\,dt \\
&= \int_0^{2\pi} \cos^2 t\,dt - \int_0^{2\pi} 2\cos^4 t\,dt + \int_0^{2\pi} \cos^6 t\,dt \\
&= \int_0^{2\pi} \cos^2 t\,dt - \int_0^{2\pi} 2\cos^4 t\,dt + \int_0^{2\pi} \cos^6 t\,dt \\
&= 2\pi\left(\frac12\right) - 2\left(2\pi\left(\frac{1\cdot3}{2\cdot4}\right)\right)
    + 2\pi\left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right) \\
&= 2\pi\left(\frac{24 - 2(18) + 15}{48}\right) \\
&= \frac18 \pi.
\end{align}
A: If I understand correctly, you are looking for a statement along the lines of:
$$ \int_0^{2\pi}  \sin^n(t)\cos^m(t) \mathrm{d}t =0, $$
if $m,n\in \mathbb{N}$, $m+n$ odd. This is easily seen as follows. Let $m,n\in \mathbb{N}$, $m+n$ odd and $$I=\int_0^{2\pi} \sin^n(t)\cos^m(t) \mathrm{d}t.$$ Then, letting $t=2\pi-u$ and using addition formulae,
$$ \int_0^{2\pi} \sin^n(t)\cos^m(t) \mathrm{d}t = -\int_{2\pi}^0\sin^n(2\pi-u) \cos^m(2\pi-u) \mathrm{du} = \int_0^{2\pi} \underbrace{(-1)^{n+m}}_{=-1} \sin^n(u)\cos^m(u)\mathrm{d}u $$
Hence, $I=-I$, which implies $I=0$.
The case where $n+m$ even means that both $m,n$ are even or odd. In this case one may efficiently use the "power-reduction" formulae. A general formula could be derived using the identities for $\sin^n(\theta)$ and $\cos^n(\theta)$ given on the linked page, but I must admit I have never seen it and I'm not sure it is in any way useful. One can certainly always calculate the integral in a closed form for given values of $m$ and $n$, e.g. $I(n=2,m=0)=I(n=0,m=2)=\pi$. I hope this answers your question.
EDIT: I incorrectly stated that in the case $m+n$ even, the integral would always be non-zero. I agree that $m=n=1$ is an obvious counter-example. In fact, the integral is always zero if at least one of $m,n$ is odd. Only if they are both even, the integral is non-zero as is shown nicely in David K's post.
