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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

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