integrate $\int{\sin(x)\cos(2x)dx}$ I can't find a way to do u-substitution for the following integral:
$$\int{\sin(x)\cos(2x)dx}$$
Is it possible to evaluate this integral?
What method(s) should I use?
 A: Hint
$$\cos(a)\sin(b) = \frac{1}{2}(\sin(a + b) - \sin(a - b))$$
A: Write
$$\cos{2 x} = 2 \cos^2{x}-1$$
Then the integral is equal to
$$-\int d(\cos{x}) (2 \cos^2{x}-1) = -\frac{2}{3}\cos^3{x} + \cos{x} + C$$
A: If you have an integral of the form: $$\int\sin nx\cos mxdx$$ Then the integral is equivalent to:$$-\frac{1}{2}\left(\frac{\cos(m+n)x}{m+n}+\frac{\cos(m-n)x}{m-n}\right)+C$$ This is very useful if you are competing in an integration bee. Anyways, substituting in $n=1$ and $m=2$ you get:$$\int\sin x\cos 2xdx=-\frac{1}{2}\left(\frac{\cos3x}{3}+\cos x\right)+C$$To prove the general formula, we could apply one of the product to sum formulas: $$\sin a\cos b=\frac{\sin(a+b)+\sin(a-b)}{2}$$This is an important formula you should memorize. Now, applying it to the integrand, we get $$\int\sin nx\cos mxdx=\int\frac{\sin((n+m)x)+\sin((n-m)x)}{2}dx$$Apply linearity: $$\int\frac{\sin((n+m)x)+\sin((n-m)x)}{2}dx=\int\frac{\sin((n+m)x)}{2}dx+\int\frac{\sin((n-m)x)}{2}dx$$By making a $u$ substitution of $(n+m)x$ and $(n-m)x$ to each integral respectively, we get the final result: $$\int\frac{\sin((n+m)x)}{2}dx+\int\frac{\sin((n-m)x)}{2}dx=-\frac{1}{2}\left(\frac{\cos(m+n)x}{m+n}+\frac{\cos(m-n)x}{m-n}\right)+C$$
A: The answer by user63181 is a good hint, but incorrect (has been fixed). The correct identity:
$$ \sin(a)\cos(b) = \frac{1}{2} (\sin(a+b) + \sin(a−b)) $$
With this identity, $\int{\sin(x)\cos(2x)dx}$ turns into
$$\frac{1}{2}\left[\int{\sin(3x) dx} + \int{\sin(-x) dx}\right]$$
Which gives
$$\int{\sin(x)\cos(2x) dx} = -\frac{1}{6} \cos(3x) + \frac{1}{2} \cos(x) + C$$
