i need some help on how to split integrals. for instance this
$\displaystyle \int_{0}^{\frac{3\pi}{2}} cosx d(|cos2x| $
thanks
If I'm not mistaken you mean you have the differential
$\displaystyle d(|cos(2x)|) = |-2 sin(2x)| dx = 2|sin(2x)| dx$
Where $\displaystyle |sin(2x)| = \begin{cases} sin(2x) & 0 \leq x < \frac{\pi}{2} \\ -sin(2x) & \frac{\pi}{2} \leq x < \pi \end{cases}$ ...(for $\displaystyle 0 \leq x < \pi$).
-Dan
You have
$\displaystyle \int_0^{3 \pi/2}cos(x)~d(|cos(2x)|) = \int_0^{3 \pi/2} cos(x) \cdot 2|sin(2x)| dx$
$\displaystyle = \int_0^{\pi /2} 2 sin(2x)~cos(x)~dx + \int_{\pi/2}^{\pi}2 cos(x)~(-sin(2x))~dx + \int_{\pi}^{3 \pi/2}{...}$
How do you do the last integral? Is sin(2x) positive or negative?
-Dan