How to find the integral $\int_0^{70 \pi} |\cos^{2}x\sin x|\,dx$? I need help with this problem:
$$\int_0^{70 \pi} \left|\cos^{2}\!\left(x\right)\sin\!\left(x\right)\right| dx$$
My friend says it's 140/3 but I don't see how.
 A: If you take the integral from $0$ to $\pi$ and then multiply by $70$, you've got it, since it's periodic with period $\pi$.  On the interval from $0$ to $\pi$ you can drop the absolute value since the function is nonnegative there.  And
$$
\int_0^{\pi/2} (\cos^2 x) \Big(\sin x\,dx\Big) = \int_1^0 u^2 \Big(-du\Big).
$$
And finally, $\displaystyle\int_{\pi/2}^\pi \cdots$ is the same thing because of geometric symmetry.  The bottom line is that your friend is right.
A way of looking at the aforementioned geometric symmetry is this:
$$
\begin{align}
\int_{\pi/2}^\pi \cos^2 x \sin x \, dx & = \int_{\pi/2}^0 \cos^2(\pi-w) \sin (\pi-w) \, \Big(-dw\Big) \\[10pt]
& = \int_0^{\pi/2} \cos^2 w \sin w\,dw  \\[10pt]
& = \int_0^{\pi/2} \cos^2 x \sin x \, dx.
\end{align}
$$
Here we used the trigonometric identities
\begin{align}
\sin(\pi - w ) & = \sin w, \\[10pt]
\text{and } \cos(\pi-w) & = - \cos w, \\
\text{whence } \cos^2(\pi-w) & = \cos^2 w.
\end{align}
A: By simetry, then
$$\int_0^{70 \pi} \left|\cos^{2}\!\left(x\right)\sin\!\left(x\right)\right| dx=70\cdot\int_0^{ \pi} \cos^{2}\!\left(x\right)\sin\!\left(x\right) dx$$
A: Hint:   $\int_0^{70\pi}=\sum_{k=0}^{34}\int_{2k\pi}^{2k\pi+2\pi}$.And
$$\int_{2k\pi}^{2k\pi+2\pi}|\cos^2(x)\sin(x)|dx=\int_{2k\pi}^{2k\pi+\pi}\sin(x)\cos^2(x)dx-\int_{2k\pi+\pi}^{2k\pi+2\pi}\sin(x)\cos^2(x)dx$$
and 
$$\sin(x)\cos^2(x)=-\dfrac{1}{3}(\cos^3(x))'$$
