how do I take the intigral of cos^2 x dx?

Answer 1

You know the derivative of sin is cos, and the integral of u^2 is (u^3)/3, so the integral must be sin^3(x)/3 + C

Answer 2

Integration Of Cos 2 X

Answer 3

Use the identity of cos^2 x = 1/2 + 1/2 cos 2x

Answer 4

Integral { sin^3(x) cos^2(x) dx U substitution and trig identities First break the sin ^3(x) into two Integral { sin x sin ^2(x) cos^2(x) dx Sin ^2(x) = 1-cos^2(x) Integral { sin x (1-cos^2(x) )(cos^2(x)) dx Now use u substitution u=cos x , du= -sin x dx, -du=sin x dx Plug in u Integral -{ (1-u^2)(u^2) du Integral -{ u^2 - u^4 du Now integrate -{ u^2 - u^4 du Integral -{ u^2 - u^4 du = -[1/3 u ^3 - 1/5 u^5 ] Plug back in u =cos x 1/5 cos ^5(x) - 1/3 cos^3(x) + C Cheers

Answer 5

So you want to solve

Integral (cos^2(x))dx

There's no way you can solve for this directly; you have to take advantage of the half angle identity, which goes as follows:

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

What you would do is take the integral of this identity that cos^2(x) is equal to. So we have:

Integral ( [1 + cos(2x)]/2)dx

Your first step is to pull out all constants. In this case, since we have a denominator of a constant, we pull out the fraction 1/2 out of the integral. This is completely a valid step, and is always a good idea to get them out of the way early.

(1/2) * Integral (1 + cos(2x))dx

At this point, solving for the integral of cos(2x) is easy; it's LIKE sin(2x), except for the fact that taking the derivative would result in the chain rule multiplying by 2, so we offset this by multiplying 1/2 for the integral.

(1/2) * [x + (1/2)sin(2x)] + C

After some steps of simplification, your final answer should be:

(1/4) [2x + sin(2x)] + C

NOTE: Contrary to one might believe, you cannot treat this like the integral of x^2, even though it is [cos(x)]^2. Integrals work differently than derivatives in that there isn't a "reverse chain rule". Intuitively, one might think that the integral is
(1/3)[cos^3(x)] + C. However, this is NOT the case, and here's the reason why.

f(x) = (1/3)[cos^3(x)] + C

Using the power rule, then the chain rule,

f'(x) = [cos^2(x)] [-sin(x)] = -sin(x)cos^2(x),

Which, as you can see, is NOT cos^2(x).
Make sure you don't fall for this fallacy of a "reverse power rule" for integrals. Integration isn't as straightforward as derivative.

Answer 6

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RE:
how do I take the intigral of cos^2 x dx?

Answer 7

Use the equality cos(2x) = 2cos²x - 1
---> cos²x = [cos(2x)+1] / 2.

This integrates to ½[½sin(2x) + x] + C = sin(2x)/4 + x/2 + C.

Answer 8

is it (cosx)^2 dx ?
if it is...then u can change it into (cosx)^2 = (1+cos(2x))/2
[from cos2x=2(cosx)^2-1]
then int (cosx)^2 dx = int (1+cos(2x))/2 dx
= (1/2) int (1+cos(2x)) dx
= (1/2) [x+(1/2)sin(2x)] + C

Answer 9

this is the real answer (sin(x)cos(x))/2 + x/2

Answer 10

cos^2x=1/2(cos2x+1) then int=x/2+1/4sin2x+C. ok!

Answer 11

Use a trig identity.

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http://www.physicsforums.com/showthread.php?t=23105