The problem involves the integration of the expression \(\int (\sin x + 2\cos x)^3\,dx\), which falls under the subject area of calculus, specifically integral calculus.
The discussion is active, with multiple approaches being explored. Participants have offered different strategies for tackling the integral, but there is no explicit consensus on a single method. The conversation reflects a collaborative effort to clarify the problem and explore potential solutions.
Some participants mention the possibility of rewriting the integrand and changing variables, indicating a focus on finding simpler forms for integration. There are also references to identities and substitutions that may or may not be necessary, suggesting varying levels of certainty about the approaches discussed.
Not sure that it helps, but you could rewrite the original integrand as ##A\sin^3(x+\alpha)##.Helly123 said:How to do this in simpler way?
That seems to be more or less the method embarked upon.andrewkirk said: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 said:Not sure that it helps, but you could rewrite the original integrand as ##A\sin^3(x+\alpha)##.
These terms can be easily integrated using a substitution. There's no need to use the Pythagorean identity here.andrewkirk said: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.