Integral of a modulus function vs absolute value of definite integral

integral of floor of x from 0 to 4 (greatest integer function)
integral of floor of x from 0 to 4 (greatest integer function)

Please refer to the diagram below.

The diagram shows a curve with equation $y = cos {x\over 2} cos x$, for 0 $\le x$ $\le$ $\pi$, along with the $x$ and $y$-intercepts of the graph.

Question: By first finding $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$, explain why $\vert \int_{\pi\over 3}^{2\pi\over 3} cos {x\over 2} cos x \,dx \vert$ is smaller than $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$. You may refer to the graph provided for assistance.

I have found $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$ and have done so as shown:

I then start to calculate $\vert \int_{\pi\over 3}^{2\pi\over 3} cos {x\over 2} cos x \,dx \vert$ as shown below.

Now, I realise I would also end up calculating $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$ eventually which implies that $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$ = $\vert \int_{\pi\over 3}^{2\pi\over 3} cos {x\over 2} cos x \,dx \vert$? But the question has already stated that $\vert \int_{\pi\over 3}^{2\pi\over 3} cos {x\over 2} cos x \,dx \vert$ will be smaller than $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$, so I’m not really sure how to proceed from here…

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