Evaluate the integral $\int (\sin x + 2\cos x)^5 dx$ Do I have to expand manually $(\sin x + 2\cos x)^5$?
I don't think I can do integration by parts or substitution with this function.
Edit: Try to make one term, what I know is using $\sin^2 x + \cos^2 x=1$, but I think I still have to expand the polynomial.
 A: One way is to recall the angle addition identity $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$  If we suppose there exists a suitable constant $k$ and angle $\theta$ such that $$\sin x + 2 \cos x = k (\sin x \cos \theta + \cos x \sin \theta),$$ then we have $$k \cos \theta = 1, \\ k \sin \theta = 2,$$ hence $$k^2 (\cos^2 \theta + \sin^2 \theta) = 1^2 + 2^2 = 5,$$ hence $k = \sqrt{5}$ and we may take $\theta = \tan^{-1} 2$.  Then we have $$\int (\sin x + 2 \cos x)^5 \, dx = \int \left( \sqrt{5} \sin (x + \theta) \right)^5 \, dx = 5^{5/2} \int \sin^5 (x + \theta) \, dx.$$  The rest is handled by the usual methods.
A: One option is simply to expand the integrand using the Binomial Theorem like you say: The integral will be a linear combination of terms of the form $$\sin^{5 - k} x \cos^k x \,dx .$$ Since one of $k, 5 - k$ is odd, we integrate each of these terms separately using a single, standard substitution.
An alternative method is to use the angle sum identity $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$ for $\sin$ backward and rewrite the quantity $\sin x + 2 \cos x$ in the form $$A \sin (x + \phi) .$$ Then, we may rewrite the integral as
$$A^5 \int \sin^5 (x + \phi) \,dx,$$
but this can be handled with a single, standard substitution. This method is faster, since you need only integrate a single term, but rewriting the antiderivative in terms of $\sin x$ and $\cos x$ involves some work, too, so if you are unhappy with an integrand expressed in terms of $\cos (x + \phi)$), you might still prefer the first method.

 We can write $$\sin x + 2 \cos x = \sqrt{5} \sin (x + \arctan 2) .$$

