If the solve the problem


$\int \sin x \cos 2 x \sin 3 x d x$


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

$\therefore \sin 3 x \cos 2 x=\frac{\sin 5 x+\sin x}{2}$

$\therefore$ The given equation becomes

$\Rightarrow \int \frac{1}{2}(\sin 5 x-\sin x) \sin x d x$

$\Rightarrow \int \frac{1}{2}\left(\sin 5 x \sin x d x-\sin ^{2} x d x\right)$

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

$\therefore \sin 5 x \sin x=\frac{\cos 4 x-\cos 6 x}{2}$

Also $\sin ^{2} x=\frac{1-\cos 2 x}{2}$

$\therefore$ Above equation can be written as

$\Rightarrow \int \frac{1}{2}\left(\frac{1}{2}(\cos 4 x-\cos 6 x) d x-\frac{1}{2}(1-\cos 2 x) d x\right)$

$\Rightarrow \frac{1}{4} \int \cos 4 x d x-\int \cos 6 x d x-\int d x+\int \cos 2 x d x$

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

$\Rightarrow \frac{1}{4}\left(\frac{1}{4} \sin 4 x-\frac{1}{6} \sin 6 x-x+\frac{1}{2} \sin 2 x+c\right)$

$\Rightarrow \frac{1}{4}\left(\frac{3 \sin 4 x-2 \sin 6 x-12+6 \sin 2 x}{12}+C\right)$

$\Rightarrow \frac{3 \sin 4 x-2 \sin 6 x-12+6 \sin 2 x}{48}+C$

NOTE: - Whenever you are solving integral questions having trigonometric functions in the product then the first thing that should be done is convert them in the form of addition or subtraction.

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