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

[Helly123]

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?

 

[haruspex]

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]

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]

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]

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]

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.