# splitting integrals!

• Jul 10th 2013, 06:52 AM
lawochekel
splitting integrals!
i need some help on how to split integrals. for instance this

$\displaystyle \int_{0}^{\frac{3\pi}{2}} cosx d(|cos2x|$

thanks
• Jul 10th 2013, 05:34 PM
topsquark
Re: splitting integrals!
Quote:

Originally Posted by lawochekel
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
• Jul 11th 2013, 02:50 AM
lawochekel
Re: splitting integrals!
pls can you be more explicit cos i don't really get it.
• Jul 12th 2013, 07:30 AM
topsquark
Re: splitting integrals!
Quote:

Originally Posted by topsquark
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
• Jul 15th 2013, 02:45 AM
lawochekel
Re: splitting integrals!
is positive sin(2x)