How to integrate $\frac{\cos2x}{ \sin x + \sin 3x}$? I thought of using trig identities to get rid of $\cos2x$ and $\sin3x$, then use Weierstrass substitution, but I got myself into big trouble as the expression got too complicated.
Is there a simpler way to tackle this problem?
 A: First, note that
$$\sin(x) + \sin(3x) = 4 \sin(x) \cos^2(x)$$
by the uncommonly-used triple angle identity for sine and the Pythagorean identity:
$$\begin{align*}
\sin(3x) 
&= 3 \sin(x) - 4 \sin^3(x) \\
&= \sin(x) \Big( 3 - 4 \sin^2(x) \Big) \\
&= \sin(x) \Big( 3 - 4 \Big( 1 - \cos^2(x) \Big) \Big) \\
&= \sin(x) \Big( 3 - 4 + 4 \cos^2(x)  \Big) \\
&= \sin(x) \Big( 4 \cos^2(x) - 1 \Big) \\
&=  4 \sin(x) \cos^2(x) - \sin(x)
\end{align*} $$
Then
$$\mathcal{I} := \int \frac{\cos(2x)}{\sin(x) + \sin(3x)} \, dx = \frac 1 4 \int \frac{\cos(2x)}{\sin(x)\cos^2(x)} \, dx$$
Then using a double-angle formula for cosine,
$$\cos(2x) = 2 \cos^2(x) - 1$$
we get
$$\mathcal{I} = \frac 1 2 \int \csc(x) \, dx - \frac 1 4 \int \frac{1}{\sin(x) \cos^2(x)} \, dx$$
The former is well-known. The second integral may be rewritten with basic trigonometry identities (turn it into $\csc(x) \sec^2(x)$,  use Pythagoras, use linearity) and solved easily.
A: Hint: $$\frac{\cos 2x}{\sin x + \sin 3x}=\frac{\cos^2 x-\sin^2 x}{2\sin 2x\cos x}=\frac{\cos^2 x-\sin^2 x}{4\sin x\cos^2 x}=\frac{1}{4\sin x}-\frac{\sin x}{4\cos^2x}$$
A: The method using the Weierstass substitution isn't so bad.
$$t=\tan\left(\frac x2\right) \implies \begin{cases}\sin(x) = \frac{2t}{1+t^2} \\ \cos(2x) = 2\left(\frac{1-t^2}{1+t^2}\right)^2 - 1 \\ \sin(3x) = \frac{6t(1-t^2)^2-8t^3}{(1+t^2)^3}\end{cases}$$
where we use the identities
$$\begin{cases}\sin(x) = 2\sin\left(\frac x2\right)\cos\left(\frac x2\right) \\ \cos(x) = \cos^2\left(\frac x2\right) - \sin^2\left(\frac x2\right) \\ \cos(2x) = 2\cos^2(x) - 1 \\ \sin(3x) = 3\cos^2(x)\sin(x) - \sin^3(x)\end{cases}$$
Now $dt = \frac12\sec^2\left(\frac x2\right)\,dx \iff dx = \frac{2\,dt}{1+t^2}$, so
$$\begin{align}
\int \frac{\cos(2x)}{\sin(x) + \sin(3x)} \, dx &= \int \frac{2\left(\frac{1-t^2}{1+t^2}\right)^2-1}{\frac{2t}{1+t^2} + \frac{6t(1-t^2)^2-8t^3}{(1+t^2)^3}} \frac2{1+t^2} \, dt \\[1ex]
&= 2 \int \frac{2(1-t^2)^2-(1+t^2)^2}{2t(1+t^2)^2 + 6t(1-t^2)^2-8t^3} \, dt \\[1ex]
&= \frac14 \int \frac{t^4-6t^2+1}{t^5-2t^3+t} \, dt \\[1ex]
&= \frac14 \int \left(\frac1t + \frac1{(t+1)^2} - \frac1{(t-1)^2}\right) \, dt \\[1ex]
\end{align}$$
