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Matilde
Guest

Hi there!

(e^x sinx)^2

## #2 2005-10-30 04:40:39

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### 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.

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Matilde
Guest

(e^x sinx)^2= ?

## #4 2005-10-30 05:24:46

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

### Re: Mathhelp!

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

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## #5 2005-10-30 05:31:40

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### 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.

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## #6 2005-10-30 07:01:29

Matilde
Member
Registered: 2005-10-30
Posts: 6

### 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?

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## #7 2005-10-30 09:21:48

Flowers4Carlos
Member
Registered: 2005-08-25
Posts: 106

### Re: Mathhelp!

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???

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## #8 2005-11-06 22:58:59

Matilde
Member
Registered: 2005-10-30
Posts: 6

### Re: Mathhelp!

I know Im late.. but ThankYouVeryMuch!

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