Integration not area In which (general) interval is the integration of
     $$\sin x\, dx$$
Greater than the integration of
  $$\sin (2x)\,dx$$
Note :it's integration not area
 A: First, you should write it with integrals.

For which $a<b\in\mathbb R$ holds $$\int_a^b \sin(x)~dx>\int_a^b\sin(2x)~dx.$$

Next, compute the integrals and you get
$$
\int_a^b\sin(x)~dx=-\cos(b)+\cos(a)
$$
and
$$
\int_a^b\sin(2x)~dx=-\frac12\cos(2b)+\frac12\cos(2a).
$$
Hence
$$
\int_a^b\sin(x)~dx>\int_a^b\sin(2x)~dx\Leftrightarrow \cos(a)-\cos(b)>\frac{\cos(2a)-\cos(2b)}2
$$
Now you can sort the variables and get
$$
\cos(a)-\cos(b)>\frac{\cos(2a)-\cos(2b)}2\Leftrightarrow 2\cos(a)-\cos(2a)>2\cos(b)-\cos(2b).
$$
If we define $f(x)=2\cos(x)-\cos(2x)$ then we have to search for all $a<b$ such that $f(a)>f(b)$
Picture of $f$
It is sufficient to consider $b-a<2\pi$, because of the periodicity of $f$.
We compute the maximal points of $f$ within $[0,2\pi]$.
$f'(x)=-2\sin(x)+2\sin(2x)=2(2\sin(x)\cos(x)-\sin(x))=2\sin(x)(2\cos(x)-1)$
We get
$$
f'(0)=0\Leftrightarrow \sin(x)=0\vee \cos(x)=\frac12\Leftrightarrow x\in\left\{0,\pi,2\pi,\frac{\pi}3,\frac{5\pi}3\right\}
$$
Here $0$,$\pi$, $2\pi$ are minimum and $\frac{\pi}3$,$\frac{5\pi}3$ are maximum of $f$. Since $f$ is symmetric at $\pi$, you can conclude:
If $a\in\left(\frac{\pi}3,\frac{5\pi}3\right)$, then $f(a)>f(b)$ holds for $b\in (a,2\pi-a)$.
(We can deduce, that its true for $b\in \bigcup_{k\geq 1}^\infty (a+2k\pi,2(k+1)\pi-a)$)
If $a\in\left\{\frac{\pi}3,\frac{5\pi}3\right\}$ we can deduce that $f(a)>f(b)$ for all $b\in(a,\infty)\setminus(\frac{\pi}3\mathbb N \cup \frac{5\pi}3\mathbb N )$.
The tricky part is for $a\in\left[0,\frac{\pi}3\right)\cup\left(\frac{5\pi}3,2\pi\right)$.
Finally you can shift the argument, if $a$ is outside of $[0,2\pi]$, the possible $b$ will just shift.
