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

## Homework Statement

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

## 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?

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haruspex
Homework Helper
Gold Member
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)##.

andrewkirk
Homework Helper
Gold Member
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.

haruspex
Homework Helper
Gold Member
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.

Ray Vickson
Homework Helper
Dearly Missed
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.

vela
Staff Emeritus