Question

integrate cos 2x cos 3x

Answer 1

hope this helps you......

Answer 2

Answer:

$\displaystyle\int\;\cos{2x}\cos{3x}\;dx=\dfrac{1}{2}\left[\dfrac{\sin{5x}}{5}-\sin{x}\right]+c$

Step-by-step explanation:

We have he given,

Let $I=\displaystyle\int\;\cos{2x}\cos{3x}\;dx\\I=\dfrac{1}{2}\displaystyle\int\;2\cos{2x}\cos{3x}\;dx$

we know that:

$2\cos a\cos b=\cos{a+b}+\cos{a-b}$

So, we have

$I=\dfrac{1}{2}\displaystyle\int\;(\cos{2x+3x}+\cos{2x-3x})\;dx\\I=\dfrac{1}{2}\displaystyle\int\;(\cos{5x}+cos{-x})\;dx\\I=\dfrac{1}{2}\displaystyle\int\;(\cos{5x}-\cos{x})\\I=\dfrac{1}{2}\left[\dfrac{\sin{5x}}{5}-\sin{x}\right]+c$