I am having trouble with this one. e^4x - (cosx)^2

:confused: We just started doing these and im lost lol

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- Apr 23rd 2007, 10:02 AMdrainAnti-derivative
I am having trouble with this one. e^4x - (cosx)^2

:confused: We just started doing these and im lost lol - Apr 23rd 2007, 10:06 AMThePerfectHacker
- Apr 24th 2007, 04:39 PMdrain
Ah thanks for that identity.

So I came up with

F = 1/4 (e^(4x)) - 1/2 * x - 1/2 (sinx) + C

:o - Apr 24th 2007, 04:48 PMJhevon
int{e^4x - (cosx)^2}dx = int{e^4x - (1 + cos2x)/2}dx

.................................= int{e^4x}dx - (1/2)int{1 + cos2x}dx

.................................= (1/4)e^4x - (1/2)x - (1/4)sin(2x) + C

by rite, you should do the integrals of e^4x and cos(2x) by substitution, but they're easy enough to do in your head. apparently you had problems with integrating cos2x. do you think you have it now? i integrated as normal, and then DIVIDED by the derivative of 2x, which is what the substitution method would amount to in this case - Apr 24th 2007, 04:50 PMdrain
Gah... I messed up the identity. Thanks.

so its (cosx)^2 = (1+cos2x)/2 ? I am bad at trig unfortunately... if you wouldn't mind explaining that identity to me as well that'd be very helpful.

I found an epxlanation online. But thank you! - Apr 24th 2007, 05:06 PMJhevon
the identity comes directly from the half angle formula, where we plug in 2x for x. do you know the half angle formula?

anyway, we can get it without the half angle formula

recall: cos(2x) = (cosx)^2 - (sinx)^2 ............we can change the (sinx)^2

....................= (cosx)^2 - (1 - (cosx)^2) = 2(cosx)^2 - 1

or we can change the (cosx)^2:

so, cos(2x) = (1 - (sinx)^2) - (sinx)^2 = 1 - 2(sinx)^2

now, if we take cos(2x) = 2(cosx)^2 - 1 and solve for (cosx)^2 we get:

cos(2x) = 2(cosx)^2 - 1

=> 1 + cos(2x) = 2(cosx)^2

=> (1 + cos(2x))/2 = (cosx)^2

if we take cos(2x) = 1 - 2(sinx)^2 and solve for (sinx)^2 we get:

cos(2x) = 1 - 2(sinx)^2

=> cos(2x) - 1 = -2(sinx)^2

=> 1 - cos(2x) = 2(sinx)^2

=> (1 - cos(2x))/2 = (sinx)^2