Definite integral of the product of powers of the sine and cosine

Question

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..

Answer 1

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), $$ \int_0^{2\pi} \sin^m t \cos^n t\,dt = 0. $$

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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}

Answer 2

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.