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#1 2005-10-31 03:37:01

Matilde
Guest

Mathhelp!

Hi there!
Can somebody please help me?

(e^x sinx)^2

#2 2005-10-31 03:40:39

mathsyperson
Moderator

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Re: Mathhelp!

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?


Why did the vector cross the road?
It wanted to be normal.

#3 2005-10-31 03:57:36

Matilde
Guest

Re: Mathhelp!

(e^x sinx)^2= ?

#4 2005-10-31 04:24:46

kylekatarn
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Re: Mathhelp!

(e^x sinx)^2 = (e^(2x)).(sinx)^2

#5 2005-10-31 04:31:40

mathsyperson
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Re: Mathhelp!

I agree. Unless there's a trigonometric identity that I am not aware of, that's the furthest it can be simplified.


Why did the vector cross the road?
It wanted to be normal.

#6 2005-10-31 06:01:29

Matilde
Novice

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Re: Mathhelp!

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?

#7 2005-10-31 08:21:48

Flowers4Carlos
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Re: Mathhelp!

hi yaz matilde!!

oh geeez.... this is a tuff one!!  i will still give it a try!!!

∫e^(2x)sinxdx                            sinx = 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 wink  who knowns???

#8 2005-11-07 21:58:59

Matilde
Novice

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Re: Mathhelp!

I know Im late.. but ThankYouVeryMuch!

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