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Topic review (newest first)
- Matilde
- 2005-11-07 21:58:59
I know I’m late.. but… ThankYouVeryMuch!
- Flowers4Carlos
- 2005-10-31 08:21:48
hi yaz matilde!!
oh geeez.... this is a tuff one!! i will still give it a try!!!
∫e^(2x)sin²xdx sin²x = 1/2(1 - cos2x)
∫e^(2x)[1/2(1 - cos2x)]dx let u=2x so du=2dx
1/2∫e^(u)[1/2(1 - cosu)]du
1/4∫e^(u)(1 - cosu)du
1/4∫e^(u)du - 1/4∫e^(u)cosudu
-1/4(∫e^(u)cosudu - ∫e^(u)du) use integration by parts on ∫e^(u)cosudu
f(u) = e^(u) g'(u) = cosu
f'(u) = e^(u) g(u) = sinu
∫f(u)g'(u)du = f(u)g(u) - ∫g(u)f'(u)du
-1/4[(e^(u)sinu - ∫e^(u)sinudu) - ∫e^(u)du] use integration by parts on ∫e^(u)sinudu
f(u) = e^(u) g'(u) = sinu
f'(u) = e^(u) g(u) = -cosu
-1/4[{e^(u)sinu - (-e^(u)cosu + ∫e^(u)cosudu)} - ∫e^(u)du]
-1/4[(e^(u)sinu + e^(u)cosu - ∫e^(u)cosudu) - ∫e^(u)du]
observe that ∫e^(u)cosudu = e^(u)sinu + e^(u)cosu - ∫e^(u)cosudu so
∫e^(u)cosudu + ∫e^(u)cosudu = e^(u)sinu + e^(u)cosu
2∫e^(u)cosudu = e^(u)sinu + e^(u)cosu
∫e^(u)cosudu = 1/2(e^(u)sinu + e^(u)cosu)
finally...
-1/4{[1/2(e^(u)sinu + e^(u)cosu)] - e^(u)}
you can plug u=2x back in there if ya want. i may have done it all wrong or there may be an easier way of doing it 
who knowns???
- Matilde
- 2005-10-31 06:01:29
Thanx kylekatarn!
I have to solve:
Volume= ∫ π (e^2x * (sinx)^2 * dx ,0 ,π
I know that the integral of e^2x = 1/2 e^2x.... but I don't know what to do with (sinx)^2?
- mathsyperson
- 2005-10-31 04:31:40
I agree. Unless there's a trigonometric identity that I am not aware of, that's the furthest it can be simplified.
- kylekatarn
- 2005-10-31 04:24:46
(e^x sinx)^2 = (e^(2x)).(sinx)^2
- Matilde
- 2005-10-31 03:57:36
(e^x sinx)^2= ?
- mathsyperson
- 2005-10-31 03:40:39
Help how? You've just typed an expression involving rather complicated algebra and it is now just sitting there looking confused and asking for its mummy. What exactly do you want done to it?
- Matilde
- 2005-10-31 03:37:01
Hi there!
Can somebody please help me?
(e^x sinx)^2