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

I think you need to use a double angle formula but I am not sure how I am supposed to split the $3x$.

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Use

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

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    $\begingroup$ Why did someone down vote this. this is perfect $\endgroup$ – Saketh Malyala Apr 10 '17 at 0:50
  • $\begingroup$ Don't know. I saw someone commenting the same thing but with a mistake, then deleting it, maybe he was angry. $\endgroup$ – Edouard Berthe Apr 10 '17 at 0:50
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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.

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    $\begingroup$ For the sake of simplicity and minimal formulas supplied on midterms I tend to keep like with like, but thank you for the insight. $\endgroup$ – goldenlinx Apr 10 '17 at 1:36
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    $\begingroup$ @goldenlinx no problem. If you go onto do Fourier analysis, the above technique will prove useful, so I thought I would offer it here to you and others. Cheers! $\endgroup$ – qbert Apr 10 '17 at 1:37
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    $\begingroup$ I'm only in first year of my mathematics degree so I will definetley be seeing that next year. Thanks again! $\endgroup$ – goldenlinx Apr 10 '17 at 1:38
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    $\begingroup$ goldie, this (Euler's formula) is actually the simplest and bestest single thing to remember. i would commit more effort remembering this than a zillion weird trig identities. $\endgroup$ – robert bristow-johnson Apr 10 '17 at 5:28
  • $\begingroup$ @robertbristow-johnson I know that this formula is widely applied beyond first year, although with the formula sheets and the vast majority of problems I face have solutions in which trigonometry is the method we are taught to use to solve them. I think that beyond the current calculus course I am in it will prove much more worth my time to invest in it rather than now for the 2 or 3 trigonometric problems I will encounter. Regardless, thank you! $\endgroup$ – goldenlinx Apr 10 '17 at 5:55
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Addition formula $$ \sin a \cos b = \frac{1}{2} \left(\sin (a+b) + \sin ( a - b)\right) \tag{1} $$


Recast integrand

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


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

integrand[1]

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    $\begingroup$ I'm not sure why you got down voted before, but this was exactly what I was looking for. Thanks! $\endgroup$ – goldenlinx Apr 10 '17 at 2:33
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Using substitution $t=x-\pi$ we get $$\int_{-\pi}^\pi \sin 2t \cos(3t+\pi) \mathrm{d}t = - \int_{-\pi}^\pi \sin 2t \cos 3t \mathrm{d}t.$$ So we are integrating and odd function and the interval is symmetric w.r.t. origin, therefore the integral is zero.

This is basically the same thing as saying that half of the Fourier coefficients of an even function are zero. (The even function here being $\cos 3t$.)

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