Integration Help

[Shaun.P]

Hi all,

I have a maths test on Tuesday at University, and I have a practice test. One of the questions I cannot find the solution to in my notes, it is:

Integrate (Cos[2x])^3, that is, Cos2x all cubed.

I don't see how to go about this problem and it is really starting to bug me! My notes only have examples for stuff like 'Sinx^3' integrate, but not with the '2x' just x.

Any help is greatly appreciated. Apparently the answer should be:

1/24 (9Sin2x + Sin6x).

 

[mlw1235]

You have to use a power reducing formula (see this page

Applying the cos^3 rule you get

.25 Int(3cos(2x) + cos(6x))

split up that integral setting u = 2x and u=6x

You the get

(3/8)sin2x - (1/24)sin6x which is equivalent to your answer.

 

[CanadaRAM]

My son keeps coming to me with math problems in his homework (in college doing University prep) and I can't remember these ^#**ers that I supposedly learned 30 years ago. Thanks for the link, it helps Dads too..

 

[iGav]

(3/8)sin2x - (1/24)sin6x which is equivalent to your answer.

what the hell does it mean???? (keep in mind you're talking to someone that doesn't know all their times tables. )

 

[Macaddicttt]

You have to use a power reducing formula (see this page

Applying the cos^3 rule you get

.25 Int(3cos(2x) + cos(6x))

split up that integral setting u = 2x and u=6x

You the get

(3/8)sin2x - (1/24)sin6x which is equivalent to your answer.

I wouldn't recommend that method since on a test you aren't going to be able to remember every power reduction formula. I suggest using the double angle formulas and 1 - (sinx)^2 = (cosx)^2 and 1 - (cosx)^2 = (sinx)^2 and just playing around with it until you come to something you can integrate.

 

[Shaun.P]

No there should be an easier way to do it.

Is there a substitution method?

I have tried writing (Cos2x)^2 x Cos2x, and tried using integrating by parts, but I don't think this is correct.

The paper is out of 30, and this question is out of 4.

I am also stuck on two other questions:

(ii) Prove the following statement in which m is an integer.

m odd => 4|(m^2 + 3) (3 marks)

I have written: If m is odd, then has forumula m = 2k + 1. m^2 + 3 = 4k^2 + 4k + 4. But I don't know how to finish this proof - Do I take out the 4 as a common factor, hence 4 (K^2 + K + 1) but how do I actually prove 4 is divisible into this?

Also,

Consider the following statement in which x is a postive real number. (3 marks)

x^(1/2) (i.e. root x) => x rational

Write down the converse of this and the contrapositive. Give a counter example to show that the converse of above is false.

Any help with any of these questions is greatly appreciated. I have all my notes - I haven't missed lectures but the examples in class are nothing like the questions asked for in the test.

 

[Shaun.P]

what the hell does it mean???? (keep in mind you're talking to someone that doesn't know all their times tables. )

I don't actually know what an integral means, other than it is an anti-derivative. If you integrate with limits, you can find the maximum area under a curve. Differentiation on the other hand is rate of change.

 

[mlw1235]

(ii) Prove the following statement in which m is an integer.

m odd => 4|(m^2 + 3) (3 marks)

I have written: If m is odd, then has forumula m = 2k + 1. m^2 + 3 = 4k^2 + 4k + 4. But I don't know how to finish this proof - Do I take out the 4 as a common factor, hence 4 (K^2 + K + 1) but how do I actually prove 4 is divisible into this?

This is quick, but I have to take a test quick here but.

m is odd.
Let m = 2a + 1 (def of odd number)
Let there be a c such that c = a^2 + a + 1 (you'll see why...)
4c = 4a^2 + 4a + 4
4c = 4a^2 + 4a + 1 + 3
4c = (2a+1)^2 + 3
4c = m^2 + 3
4 | m^2 + 3


fyi....start at the top of the proof, and go as far as you can, then go to the bottom and work your way up.