Subjective Type

Find the integrals of the functions $$\sin x \sin 2x \sin 3x$$

Solution

It is known that, $$\sin A \sin B=\frac {1}{2}\left \{\cos (A-B)-\cos (A+B)\right \}$$
$$\therefore\displaystyle \int \sin x \sin 2x \sin 3x dx=\int [\sin x\cdot \frac {1}{2}\left \{\cos (2x-3x)-\cos (2x+3x)\right \}]dx$$
$$\displaystyle =\frac {1}{2}\int (\sin x \cos (-x)-\sin x \cos 5x)dx$$
$$\displaystyle =\frac {1}{2}\int (\sin x \cos x-\sin x \cos 5x)dx$$
$$\displaystyle =\frac {1}{2}\int \frac {\sin 2x}{2}dx-\frac {1}{2}\int \sin x \cos 5x dx$$
$$\displaystyle =\frac {1}{4}\left [\frac {-\cos 2x}{2}\right ]-\frac {1}{2}\int \left \{\frac {1}{2}\sin (x+5x)+\sin (x-5x)\right \}dx$$
$$\displaystyle =\frac {-\cos 2x}{8}-\frac {1}{4}\int (\sin 6x+\sin (-4x))dx$$
$$\displaystyle =\frac {-\cos 2x}{8}-\frac {1}{4}\left [\frac {-\cos 6x}{3}+\frac {\cos 4x}{4}\right ]+C$$
$$\displaystyle =\frac {-\cos 2x}{8}-\frac {1}{8}\left [\frac {-\cos 6x}{3}+\frac {\cos 4x}{2}\right ]+C$$
$$\displaystyle =\frac {1}{8}\left [\frac {\cos 6x}{3}-\frac {\cos 4x}{2}-\cos 2x\right ]+C$$

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