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

  • #1
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Homework Statement



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

Homework Equations




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?
 

Answers and Replies

  • #2
haruspex
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How to do this in simpler way?
Not sure that it helps, but you could rewrite the original integrand as ##A\sin^3(x+\alpha)##.
 
  • #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.
 
  • #4
haruspex
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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.
That seems to be more or less the method embarked upon.
 
  • #5
Ray Vickson
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Not sure that it helps, but you could rewrite the original integrand as ##A\sin^3(x+\alpha)##.
Then changing the integration variable to ##u = x + \alpha## would simplify the task even more.
 
  • #6
vela
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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.
These terms can be easily integrated using a substitution. There's no need to use the Pythagorean identity here.
 

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