I have no idea how to do this one, they intercept at some random point that I can not calculate.
$y= \cos x$, $y=\sin 2x$, $x= 0$ and $x= \pi/2$
I know the graph will go from $0$ to $0$ for $\sin 2x$ and then $1$ to $0$ for $\cos x$ so there are two areas I have to compute but I have no idea how to figure out the bounds on them.
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$\begingroup$Rewriting $$\sin 2x = \sin x \cos x + \cos x \sin x = 2\sin x\cos x$$ we can compute the intersection: $\cos x = \sin(2x)$ is the same as $$\begin{align*} \cos x&= 2\sin x\cos x\\ \cos x - 2\sin x\cos x &= 0\\ \cos x(1 - 2\sin x) &= 0. \end{align*}$$ The product is zero if and only if $\cos x = 0$ (which on $[0,\pi/2]$ occurs only at $x=\pi/2$), or if $1-2\sin x = 0$, which is the same as $2\sin x = 1$, which is the same as $\sin x = \frac{1}{2}$; on $[0,\pi/2]$, this happens once and only once: at $x=\pi/6$.
So the point of intersection is at $x=\pi/6$.
On $[0,\pi/6]$, we have that $\cos(x)$ is greater than $\sin(2x)$. On $[\pi/6,\pi/2]$, we have that $\sin(2x)$ is greater than $\cos x$. So the area is given by $$\begin{align*} \text{Area} &= \int_0^{\pi/2}|\cos x-\sin(2x)|\,dx\\ &= \int_0^{\pi/6}|\cos x - \sin(2x)|\,dx + \int_{\pi/6}^{\pi/2} |\cos x - \sin(2x)|\,dx\\ &= \int_0^{\pi/6}(\cos x - \sin (2x))\,dx + \int_{\pi/6}^{\pi/2}(\sin(2x) - \cos x)\,dx. \end{align*}$$ Now you can simply compute the integrals and add up the appropriate quantities.
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