A problem in integration. As you know from basic trigonometry that $\sin(2x) = 2\sin(x)\cos(x)$. If you integrate both sides with respect to x, one finds $$\int \sin(2x) \ dx =  -\frac{1}{2}\cos(2x)+c$$ on the left hand side and $$2\int\sin(x)\cos(x)\  dx = \sin^2(x)+c$$ for the right hand side.
They are different so what is the true integral?  
 A: Since, as it was pointed out, $\cos(2x) = 1-2\sin^2 x$, you can see that
$$\sin^2(x) + c = -\frac12 \cos(2x) + (\frac12 + c)$$
This means that both integrals describe the same set of solutions. Remember that anytime you are integrating a function $f$, you get a set of functions for which $F'=f$. The set always contains functions that differ by only a constant, so you can take any single function out of the set and use it do describe the whole result.

You said you want to also integrate the functions from $0$ to $\frac\pi6$. Well, from what we already know, you can be sure that the result will be the same. Why? Well, if $F_1(x) = F_2(x) + C$, and $C$ is a constant, and we know that $F_1'=F_2'=f$, then we already know that
$$\int_a^b f(x)dx = F_1(b)-F_1(a) = F_1(b) - F_1(a) + C - C = \\=F_1(b) + c - (F_2(b) + C) = F_2(b) - F_2(a)$$
So the integral must be the same no matter which function you take.
In your particular example, you get
$$\int_0^\frac\pi6 \sin(2x) = \left.-\frac12\cos(2x)\right|_0^{\frac\pi6} = -\frac12\cos\left(\frac\pi3\right) + \frac12\cos 0 = -\frac14+\frac12=\frac14$$
and
$$2\int_0^\frac\pi6 \sin x\cos x = \left.\sin^2 x\right|_0^\frac\pi6 = \sin^2\left(\frac\pi6\right) - \sin^2 0 = \left(\frac12\right)^2 - 0 = \frac14$$
A: Hint: recall that
$$
\cos (2x) = 1-2\sin^2 x
$$
