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

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.

• why is there a minus sign between your fraction and the cos term? Secondly, cosine of 0 is not 0 – imranfat Sep 7 '13 at 20:44

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