# Anti-derivative

• Apr 23rd 2007, 10:02 AM
drain
Anti-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 AM
ThePerfectHacker
Quote:

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

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

The antiderivative of e^{4x} is (1/4)e^{4x} +C if you use the substitution t=4x.

The antiderivative of (cos x)^2 can be simplified if you use the half-angle identity,

(cos x)^2 = (1+ cos (x))/2
• Apr 24th 2007, 04:39 PM
drain
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 PM
Jhevon
Quote:

Originally Posted by drain
Ah thanks for that identity.

So I came up with

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

:o

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 PM
drain
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 PM
Jhevon
Quote:

Originally Posted by drain
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.

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