The value of the definite integral int_{0}^{f | vedclass

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(D) Let $I = \int_{0}^{\frac{\pi}{2}} \sin x \sin 2x \sin 3x \, dx$. Using the trigonometric identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A

Vedclass · Accepted Answer

$\frac{1}{6}$

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(D) Let $I = \int_{0}^{\frac{\pi}{2}} \sin x \sin 2x \sin 3x \, dx$. Using the trigonometric identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$,we have $\sin x \sin 3x = \frac{1}{2} [\cos(2x) - \cos(4x)]$. Substituting this into the integral: $I = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} [\cos(2x) - \cos(4x)] \sin 2x \, dx$ $I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [\sin 2x \cos 2x - \sin 2x \cos 4x] \, dx$ Using $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$ and $\sin 2x \cos 4x = \frac{1}{2} [\sin(6x) - \sin(2x)]$: $I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [\frac{1}{2} \sin 4x - \frac{1}{2} \sin 6x + \frac{1}{2} \sin 2x] \, dx$ $I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} [\sin 4x - \sin 6x + \sin 2x] \, dx$ Integrating term by term: $I = \frac{1}{4} [-\frac{\cos 4x}{4} + \frac{\cos 6x}{6} - \frac{\cos 2x}{2}]_{0}^{\frac{\pi}{2}}$ Evaluating at limits: $I = \frac{1}{4} [(-\frac{\cos 2\pi}{4} + \frac{\cos 3\pi}{6} - \frac{\cos \pi}{2}) - (-\frac{\cos 0}{4} + \frac{\cos 0}{6} - \frac{\cos 0}{2})]$ $I = \frac{1}{4} [(-\frac{1}{4} - \frac{1}{6} + \frac{1}{2}) - (-\frac{1}{4} + \frac{1}{6} - \frac{1}{2})]$ $I = \frac{1}{4} [-\frac{1}{4} - \frac{1}{6} + \frac{1}{2} + \frac{1}{4} - \frac{1}{6} + \frac{1}{2}] = \frac{1}{4} [1 - \frac{1}{3}] = \frac{1}{4} 	imes \frac{2}{3} = \frac{1}{6}$.

The value of the definite integral $\int_{0}^{\frac{\pi}{2}} \sin x \sin 2x \sin 3x \, dx$ is equal to:

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