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

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1. Dec 7, 2018 at 11:43 PM

### Helly123

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. Dec 8, 2018 at 12:20 AM

### haruspex

Not sure that it helps, but you could rewrite the original integrand as $A\sin^3(x+\alpha)$.

3. Dec 8, 2018 at 12:27 AM

### andrewkirk

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. Dec 8, 2018 at 12:29 AM

### haruspex

That seems to be more or less the method embarked upon.

5. Dec 8, 2018 at 7:48 AM

### Ray Vickson

Then changing the integration variable to $u = x + \alpha$ would simplify the task even more.

6. Dec 8, 2018 at 11:29 PM

### vela

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