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The integral in question is this: $$ \int_{0}^{140\pi} | \sin^2(x) \cos(x)|dx $$ I tried to work this out but I don't know where to start, and I looked to WolframAlpha, but even it didn't give me an answer. Thanks for any help!

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Hint #1: That function is periodic with period $\pi$. So, your integral is$$140\times\int_0^\pi\bigl\lvert\sin^2(x)\cos(x)\bigr\rvert\,\mathrm dx.$$Hint #2: $\displaystyle\int_0^\pi\bigl\lvert\sin^2(x)\cos(x)\bigr\rvert\,\mathrm dx=\int_0^{\frac\pi2}\sin^2(x)\cos(x)\,\mathrm dx-\int_{\frac\pi2}^\pi\sin^2(x)\cos(x)\,\mathrm dx$.

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You want $$ \int_0^{140\pi} |\sin^2 x \cos x|\, dx $$ Because of the absolute value, we need to distinguish regions where the integrand is negative, so we can change the sign in those regions.

But the integrand is negative from $\frac{\pi}2 + 2n\pi$ to $3\frac{\pi}2 + 2n\pi$ and in those ranges, it twice replicates the set of values on $[0,\frac\pi2]$.
So the answer will be $$ 280 \int_0^\frac\pi2 \sin^2 x \cos x \,dx = 280 \left[ \frac13 \sin^3 x\right]_0^\frac\pi2 = \frac{280}3$$

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