triple integration

Quote:

Originally Posted by sonia1

Can anyone help plzzzz...

$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\int_{0}^{\pi/2}\cos(x+y)*cos(y+z)dx dydz$

Since we are integrating over a rectangle Fubini's theorem tells us the order of the interated integrals does not matter.

first w.r.t x we get

$\int_{0}^{\pi/2}\cos(x+y)\cos(y+z)dx=\sin(x+y)\cos(y+z) \bigg|_{0}^{\pi /2}=$

$\sin\left(\frac{\pi}{2}+y \right)\cos(y+z)-\sin(y)\cos(y+z)$

Now we need to integrate this with respect to z

$\int_{0}^{\pi/2}\sin\left(\frac{\pi}{2}+y \right)\cos(y+z)-\sin(y)\cos(y+z)dz=$

$\sin\left(\frac{\pi}{2}+y \right)\sin(y+z)-\sin(y)\sin(y+z) \bigg|_{0}^{\pi/2}=$

$\sin^{2}\left(\frac{\pi}{2}+y \right)-\sin(y)\sin\left(y+\frac{\pi}{2}\right)-\left[ \sin\left(\frac{\pi}{2}+y \right)\sin(y)-\sin(y)\sin(y)\right]$

Now finally we only have the integral w.r.t y left so we get

$\int_{0}^{\pi/2} \sin^{2}\left(\frac{\pi}{2}+y \right)-2\sin(y)\sin\left(y+\frac{\pi}{2}\right)+\sin^{2}( y)dy$

I will leave the last integral for you just use a few trig identites.

Good luck