Remembering that $\sin^2(\theta) = 1/2 - 1/2\cos(2\theta)$? How do you remember this for integrals?
It doesn't seem obvious and I can never remember it when I come across it in integrals.
 A: The double angles should be remembered as follows,
\begin{align*}
2\sin^2 (\theta) &= 1 - \cos(2\theta) \\
2\cos^2 (\theta) & = 1 + \cos(2 \theta).
\end{align*}
To know which sign for each, remember that 'sinning' is bad, hence it leads to $-$.
Reason why you should memorize like this is because there no fractions and you can see a pattern!
When you need to compute the integral of say, $\sin^2 (\theta)$, just recall formula and divide by $2$.
A: HINT:
I remember the identity by first remembering the addition formula:
$$\cos (x+y)=\cos x \cos y-\sin x \sin y \implies \cos (2x) =1-2\sin^2 x$$
A: here is a geometric way to remember the double angle formulae. we will use the fact that on the unit circle the terminal point corresponding to the arc length $t$ is $(\cos t, \sin t).$
look at the arc of length $2t$  starting at $A = (1,0)$ and ending at $B = (\cos 2t , \sin 2t)$ the midpoint $C$ of $AB$ has coordinate $((1+\cos 2t)/2, \sin 2t/2).$ but the length $OC$ is $\cos t.$ 
now the point on the unit circle cut by $OC$ can be seen in two ways: 
$$a: (\cos t, \sin t), \quad b: \left(\frac{1+\cos 2t}{2\cos t}, \frac{\sin 2t}{2\cos t}\right)$$
equation the two gives the double angle formula $$1+\cos 2t = 2\cos^2 t, \sin 2t = 2\sin t \cos t $$
A: If I want to recall the formulas
$|\sin \frac x2| = \sqrt{\frac{1-\cos x}2}$ and 
$|\cos \frac x2| = \sqrt{\frac{1+\cos x}2}$ 
and if I already remember the form $$\sqrt{\frac{1\pm\cos x}2}$$ but I do not remember which sign corresponds to sine and which to cosine, I can recall it like this:
If $x$ is close to $0$, then $\cos x$ is close to $1$. I want to get the value of $\sin(x/2)$, which is close to $0$, which I get from $1-\cos x$. Similarly, I want to get $\cos(x/2)$ close to $1$, to get this, I take $1+\cos x$.
But I agree that probably the better way is to recall how the formulas were derived from the double angle formulas.
A: Its easy to remember cos(2x) = cos2x - sin2x
and then use cos2x + sin2x = 1 to get the required identity
A: $\cos^2 \theta + \sin^2 \theta = 1$
$\cos^2 \theta - \sin^2 \theta = \cos 2\theta$  
Add and you get
$2\cos^2 \theta = 1 + \cos 2\theta$
$\cos^2 \theta = \frac 1 2 + \frac 1 2\cos 2\theta$
Subtract and you get
$2\sin^2 \theta = 1 - \cos 2\theta$
$\sin^2 \theta = \frac 1 2 - \frac 1 2\cos 2\theta$
A: Here's a longer explanation, which maybe will make the identity seem a bit more obvious:
As you know, a cosine wave oscillates between $-1$ and $+1$ with a period of $2 \pi$.
Here are some of the values:
$$
\cos 0=+1
,\quad
\cos(\pi/2) = 0
,\quad
\cos\pi = -1
,\quad
\cos(3\pi/2) = 0
,\quad
\cos(2\pi) = +1
.
$$
Compare this to the values you get if you square everything:
$$
\cos^2 0=+1
,\quad
\cos^2(\pi/2) = 0
,\quad
\cos^2\pi = +1
,\quad
\cos^2(3\pi/2) = 0
,\quad
\cos^2 2\pi = +1
.
$$
So a cosine-squared wave oscillates between $0$ and $+1$, and it makes two oscillations where the cosine wave makes only one, so it oscillates twice as fast.
In fact, it's a perfect sin/cos-wave shaped oscillation. Of course, this isn't obvious just from sampling a few values, but it's true; that's what the double-angle formula is saying. Try plotting it on a computer, and you'll see that it certainly looks that way!
So it must be oscillating around its average value $1/2$ with an amplitude of $1/2$ (so that it precisely reaches up to its maximum value $1$ and reaches down to its minimum value $0$).
Thus, you get the cosine-squared wave by taking a cosine wave $\cos 2\theta$ (with twice the frequency compared to $\cos \theta$), multiplying it by the amplitude factor $1/2$, and then adding $1/2$ to shift the graph upwards:
$$
\cos^2 2 \theta = \frac12 + \frac12 \cos 2\theta
.
$$
And the formula for the sine-squared that you asked about is exactly similar;
it's just that you multiply by the amplitude factor $-1/2$ instead, to flip the wave upside down. (Since it starts out with $\sin^2 0 = 0 = 1/2 - 1/2$ instead of $\cos^2 0 = 1 = 1/2 + 1/2$.)
