Mathematics

Find the integral of the function $$\sin 3x\cos 4x$$.


SOLUTION
for integrals of type $$\sin ax\, \cos bx$$
$$\sin ax\ \cos bx =\dfrac{1}{2}\left( \sin(ax+bx)+\sin(ax-bx) \right )$$
If $$a=3$$ $$b=4$$ 
$$\sin 3x\ \cos 4x=\dfrac{1}{2}\left( \sin(7x)+\sin(-x) \right )$$
                        $$=\dfrac{1}{2}\left( \sin 7x-\sin x \right )$$
$$\displaystyle \int \sin 3x\cos 4 x \,dx = \dfrac{1}{2}\int \left(\sin 7x-\sin x \right )dx$$ 
                                  $$=\dfrac{1}{2}\left(\dfrac{-1}{7}\cos 7x+\cos x \right )+c$$

                                   $$=\dfrac{\cos x}{2}-\dfrac{-\cos 7x}{14}+c$$ 
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Subjective Medium Published on 17th 09, 2020
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