Product of Sines and Cosines (mixed even and odd powers or only
odd powers)

In this video we show how to consider integrals of the form
$$\int \sin^n(x) \cos^m(x)\, dx.$$ There are 5 cases:

If $m=1$, we can do a u-substitution with $u=\sin(x)$ to
convert it to $\int u^n\, du$.

If $m$ is odd, we can use the identity $\cos^2(x) =
1-\sin^2(x)$ to convert it to case 1.

If $n=1$, we can do a u-substitution with $u=\cos(x)$ to
convert it to $\int -u^m\, du$.

If $n$ is odd, we can use the identity
$\sin^2(x)=1-\cos^2(x)$ to convert it to case 3.

If $m$ and $n$ are both even, we need to use double-angle
formulas (see the next page).

Alternatively,
for those of you who find it hard to remember the rules for
whether $m$ or $n$ is odd or even, it may helpful to think of
these problems as stripping off the derivative
of a trig function so that the trig function, which becomes
your $u$, is all that remains in the rest of the integrand.
Let's look at some examples.

When the power of cosine is 1, or the power of sine
is 1, then the substitutions $u=\sin x$ or $u=\cos x$
(respectively) will work, since the derivatives $\cos x\,dx$ or
$\sin x\,dx$ (respectively) will appear in the integral without
having to do anything.

Now examine $\int\sin^2(x)\cos^3(x)\,dx$. We
could strip off $\sin x\,dx$, or $\cos x\, dx$. But if we
strip off the sine, we are left with both sines and cosines:
$\int\sin x\cos^3 x\sin x\,dx$ with no easy way to fix it.
Let's strip off $\cos x\,dx$. Then after applying the
Pythagorean trig indentity, we are left with just sines:
$\int(\sin^2 x\cos^2 x\cos x\,dx=\int\sin^2(x)(1-\sin^2(x))\cos
x\,dx$ and can now happily let $u=\sin x$.

DO: Try to figure
out what derivative of what trig function to strip off in
$\int\sin^5 x\cos^4 x\,dx$ without looking at whether $m$ or $n$
is odd or even, just by trying the two alternatives: stripping
off $\sin x\,dx$ or $\cos x\,dx$.

(The answer is $\sin x\,dx$, leaving $\sin^4 x\cos^4
x=(1-\cos^2 x)^2\cos^4 x$, just cosines, so $u=\cos x$ works.)