How would you solve this integral?

Question

Indefinite Integral of (cos(7x)^4)dx

Answer 1

It's (3/8) x + (1/28) Sin(14 x) + (1/224) Sin(28 x)

Best way to do this is by using the exponential representation of trig functions, otherwise you can get lost pretty easily.

Sure, "It's not hard to do!", if you can somehow keep track of a million terms without making a mistake, doing this in the conventional trig form.

Answer 2

do no longer know what's up with that y^5 up there, yet right here is going your vital... int y^4 from 0 to 2 = y^5/5 2^5/5 - 0.5/5 32/5 - 0 vital = 32/5 i can not clarify right here the thank you to sparkling up integrals. there are various diverse techniques for many categories of integrals. Yours is the least confusing, yet explaining why and how would take a protracted, long term.

Answer 3

Integral ( cos^4(7x) dx)

First, use the property that cos^4(7x) = [cos^2(7x)]^2

Integral ( [cos^2(7x)]^2 dx)

Now, use the half angle identity. Remember that

cos^2(y) = (1/2) (1 + cos(2y))
Since we have a 7x as our "y", we're going to end up with 14x as a result.

Integral ( [(1/2) (1 + cos(14x))]^2 dx )

Now, let's square both the (1/2) and the (1 + cos(14x)).

Integral ( (1/4) [ (1 + cos(14x)) ]^2 dx )

Pull the (1/4) out as a constant.

(1/4) * Integral ( [ (1 + cos(14x)) ]^2 dx )

Square the binomial.

(1/4) * Integral ( [ 1 + 2cos(14x) + cos^2(14x)] dx )

Apply the half angle identity to cos^2(14x).

(1/4) * Integral ( [ 1 + 2cos(14x) + (1/2)(1 + cos(28x)] dx )

Distribute the (1/2), to get

(1/4) * Integral ( [ 1 + 2cos(14x) + 1/2 + (1/2)cos(28x) ] dx )

(1/4) * Integral ( [ 3/2 + 2cos(14x) + (1/2)cos(28x) ] dx )

Now, we can integrate each of them individually, noting that the integral of cos (kx) is (1/k) sin(kx).

(1/4) [ (3/2)x + 2 (1/14) (sin(14x)) + (1/2) (1/28) sin(28x) ] + C

Distributing the (1/4), to get

(3/8)x + (1/28)sin(14x) + (1/4)(1/56)sin(28x) + C

(3/8)x + (1/28)sin(14x) + (1/4)(1/56)sin(28x) + C

(3/8)x + (1/28)sin(14x) + (1/224)sin(28x) + C

Answer 4

First, break cos^2n down using power reduction formulas. Then, once you have all your trig functions to the first power, I'd then use a simple u-sub and evaluate this one.

It's not hard to do, it's hard to remember all those trig formulas!

Viola

Answer 5

How would you solve it?
I would ask for help on 'Yahoo Answers'