If the solve the problem


$\int \cos 3 x \cos 4 x d x$


We know $2 \cos A \cos B=\cos (A-B)+\cos (A+B)$

$\therefore \cos 4 x \cos 3 x=\frac{\cos x+\cos 7 x}{2}$

$\therefore$ The above equation becomes

$\Rightarrow \int \frac{1}{2}(\cos x-\cos 7 x) d x$

$\Rightarrow \frac{1}{2}\left(\int \cos x d x-\int \cos 7 x d x\right)$

We know $\int \cos a x d x=\frac{1}{a} \sin a x+c$

$\Rightarrow \frac{1}{2}\left(\sin x-\frac{1}{7} \sin 7 x\right)$

$\Rightarrow \frac{7 \sin x-\sin 7 x}{14}+C$

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