Evaluate: 4 x 3 x d x - Vidyadip

Question

Evaluate:
$\int \sin 4 x \cos 3 x d x$

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Answer

$\int \sin 4 x \cos 3 x d x$
$=\frac{1}{2} \int \sin 4 x \cos 3 x d x$
$=\frac{1}{2} \int[\sin (4 x+3 x)+\sin (4 x-3 x)] d x$
$=\frac{1}{2} \int \sin 7 x d x+\frac{1}{2} \int \sin x d x$
$=\frac{1}{2}\left(\frac{-\cos 7 x}{7}\right)-\frac{1}{2} \cos x+c$
$=-\frac{1}{14} \cos 7 x-\frac{1}{2} \cos x+c .$

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