Trigonometric Integration ${\int_{0}^{2\pi}} \sin(2x)\cos(3x)\, dx$

Use

$$\sin(a)\cos(b) = \frac{1}{2}\left(\sin(a+b) + \sin(a-b)\right)$$

If you would like to avoid any of those trig formulas and know the complex definition of $\sin$ and $\cos$ this is the easy to compute integral: $$ \frac{1}{4i}\int_0^{2\pi}(e^{2i\theta}-e^{-2i\theta})(e^{3i\theta}+e^{-3i\theta})d\theta $$ which you will find is zero by multiplying stuff and evaluating easy integrals by substitution.

Addition formula $$ \sin a \cos b = \frac{1}{2} \left(\sin (a+b) + \sin ( a - b)\right) \tag{1} $$

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Recast integrand

For the integrand, use the addition formula $(1)$ with $a=2x$ and $b=3x$ to find $$ \sin (2 x) \cos (3 x) = \frac{1}{2} \left(\sin (5 x)-\sin (x)\right) $$

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Transform integral $$ \int \sin (2 x) \cos (3 x) dx = \frac{1}{2} \int \sin (5 x)-\sin (x)\, dx = \frac{1}{2} \cos x - \frac{1}{10} \cos (5x) $$