Find the area of the region of y= cos x and y=sin2x, x= 0, x=pi/2 i sketched the curve and y=cos x is on top, so wouldn't it be \int cosx-sin2x from pi/2 to 0?
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Are you sure that $\displaystyle \cos x \geq \sin{2x}$ on $\displaystyle x\in[0,\pi/2]$? What about $\displaystyle x=\pi/4$? You need to break up the integral into intervals where one function is greater than or equal to the other.
so would I have to split it into 2 integrals, one where cos x is bigger than sin2x, and the other where sin2x is bigger than cos x?
Yes. In order to do that you need to solve $\displaystyle \cos x = \sin{2x}\Leftrightarrow \cos x = 2\cos x \sin x$
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