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How would I integrate the following

$\int_0^{\frac{\pi}{2}}\cos(2x)\sin(3x)\,dx$

I did the following

$\frac{1}{2}\int \left(\sin(5x)-\sin(x)\right)\,dx$

$\frac{1}{10}-\cos(5x)+cos(x)$

$\frac{1}{10}(0)+0-0=0$

But I am not sure If I did it right.

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  • $\begingroup$ why is there a minus sign between your fraction and the cos term? Secondly, cosine of 0 is not 0 $\endgroup$ – imranfat Sep 7 '13 at 20:44
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Not quite. Using the identity

$$2 \cos{a}\sin{b} = \sin{(a + b)} - \sin{(a - b)}$$

we find that

\begin{align} \int_0^{\pi/2} \cos{(2x)} \sin{(3x)} &= \frac 1 2 \int_0^{\pi/2} \sin{5x} - \sin{(-x)} \\ &= \frac 1 2 \int_0^{\pi/2} \sin{5x} + \sin{x} \\ &= \frac 1 2 \left(\frac{-1}{5} \cos{5x} - \cos{x}\right)_0^{\pi/2} \\ &= \frac 1 2 \left(\frac{-1}{5} \cos{\frac{5\pi}{2}} - \cos{\frac{\pi}{2}} - (\frac{-1}{5} \cos{0} - \cos{0})\right) \\ &= \frac 1 2 \left(0 - (\frac{-1}{5} - 1)\right) \\ &= \frac 3 5 \end{align}

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$$\int\limits_0^{\frac{\pi}{2}}\cos{2x}\sin{3x}\ dx=\frac{1}{2}\int\limits_0^{\frac{\pi}{2}}{(\sin{5x}+\sin{x})\ dx}=\\ =\left. \left(-\frac{1}{10}\cos{5x}-\frac{1}{2}\cos{x}\right) \right|_{0}^{\frac{\pi}{2}}=\frac{1}{10}+\frac{1}{2}$$

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