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Homework Help: ##\int (\sin x + 2\cos x)^3\,dx##

  1. Dec 7, 2018 at 11:43 PM #1
    1. The problem statement, all variables and given/known data

    $$\int (sinx + 2cos x)^3dx$$
    2. Relevant equations


    3. The attempt at a solution
    $$\int (sinx + 2cos x)^3dx$$
    $$\int (sinx + 2cos x)((sinx + 2cos x)^2dx)$$
    $$\int (sinx + 2cos x)(1 + 3cos^2x+2sin2x)dx$$
    How to do this in simpler way?
     
  2. jcsd
  3. Dec 8, 2018 at 12:20 AM #2

    haruspex

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    Not sure that it helps, but you could rewrite the original integrand as ##A\sin^3(x+\alpha)##.
     
  4. Dec 8, 2018 at 12:27 AM #3

    andrewkirk

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    First do a binomial expansion of the integrand. That will have only four terms.
    The powers of the sine and cosine in each term will either be 3 and 0, or 2 and 1 (unordered pairs). Where they are 2 and 1, use the identity that sine squared plus cos squared equals 1 to turn the squared trig function into the other sort of trig function. Then you will have six terms, each of which is a power of only sine or only cosine.
    Then use de Moivre's formula to convert the higher powers into multiple-angle sines or cosines with power 1.
    Integration will then be straightforward.
     
  5. Dec 8, 2018 at 12:29 AM #4

    haruspex

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    That seems to be more or less the method embarked upon.
     
  6. Dec 8, 2018 at 7:48 AM #5

    Ray Vickson

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    Then changing the integration variable to ##u = x + \alpha## would simplify the task even more.
     
  7. Dec 8, 2018 at 11:29 PM #6

    vela

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    These terms can be easily integrated using a substitution. There's no need to use the Pythagorean identity here.
     
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