1. ## integration

The integral (from 0 to pi/2) of (costheta)(sintheta) d(theta)

Thank you very much.

2. Originally Posted by kittycat
The integral (from 0 to pi/2) of (costheta)(sintheta) d(theta)

Thank you very much.

$\int_{0}^{\frac {\pi}{2}} \cos \theta \sin \theta ~d \theta = \frac {1}{2} \int_{0}^{\frac {\pi}{2}} \sin 2 \theta~d \theta$ which is a simple integration by substitution problem

or you can use substitution directly. let $u = \sin \theta$

hi Jhevon,

the integral (from 0 to pi/2) of (cos^2 theta) (sintheta) d(theta)

Thank you very much.

4. Originally Posted by kittycat
hi Jhevon,

the integral (from 0 to pi/2) of (cos^2 theta) (sintheta) d(theta)

Thank you very much.
substitution: $u = \cos \theta$

(remember, if you're having trouble with the integrals even after i tell you the substitutions, don't be afraid to say so)

5. 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.

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.

6. Originally Posted by kittycat
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.

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.
are you sure 8/(4^a) shouldn't be 8/(a^4) ??

7. you've gone silent on me. assuming it should have been a^4 instead of 4^a, we proceed as follows:

$\frac {8}{a^4} \int_{0}^{ \frac {\pi}{2}} \cos^2 \theta \sin \theta~d \theta \int_{0}^{a} r^4~dr = \frac {8}{a^4} \left[ - \frac {1}{3} \cos^3 \theta \right]_{0}^{ \frac {\pi}{2}} \left[ \frac {1}{5}r^5 \right]_{0}^{a}$

(as i said earlier, i used the substitution $u = \cos \theta$ for the first integral)

........................................... $= \frac {8}{a^4} \left[ \frac {1}{3} \right] \left[ \frac {a^5}{5} \right]$

........................................... $= \frac {8a}{15}$

8. You can also use the general result.

9. Hi Jhevon,

Thank you very much.

I am sorry because I encountered trouble in re-connecting my internet.

Yes, you are right. It should be 8/a^4

Thank you very much.