The integral (from 0 to pi/2) of (costheta)(sintheta) d(theta)
Thank you very much.
remember your identities
$\displaystyle \int_{0}^{\frac {\pi}{2}} \cos \theta \sin \theta ~d \theta = \frac {1}{2} \int_{0}^{\frac {\pi}{2}} \sin 2 \theta~d \theta$ which is a simple integration by substitution problem
or you can use substitution directly. let $\displaystyle u = \sin \theta $
hi Jhevon,
Thank you very much for your kindness.
Actually the question is (8/4^a) times the integral (from 0 to pi/2) of (cos^2 theta) (sintheta) d(theta) times the integral (from 0 to a) r^4 dr.
The answer is (8a)/15.
I have trouble to get to this answer. If you could help, please show me how you can arrive this answer. Thank you very much.
you've gone silent on me. assuming it should have been a^4 instead of 4^a, we proceed as follows:
$\displaystyle \frac {8}{a^4} \int_{0}^{ \frac {\pi}{2}} \cos^2 \theta \sin \theta~d \theta \int_{0}^{a} r^4~dr = \frac {8}{a^4} \left[ - \frac {1}{3} \cos^3 \theta \right]_{0}^{ \frac {\pi}{2}} \left[ \frac {1}{5}r^5 \right]_{0}^{a}$
(as i said earlier, i used the substitution $\displaystyle u = \cos \theta$ for the first integral)
...........................................$\displaystyle = \frac {8}{a^4} \left[ \frac {1}{3} \right] \left[ \frac {a^5}{5} \right]$
...........................................$\displaystyle = \frac {8a}{15}$
You can also use the general result.