Question:

To evaluate the integral $I = \int_0^{\pi/2} \sin x \sin 2x \sin 3x \,dx$: Step 1: Apply product-to-sum identities. Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$, we first simplify $\sin x \sin 3x$: $$ \sin x \sin 3x = \frac{1}{2} (\cos(x-3x) - \cos(x+3x)) = \frac{1}{2} (\cos(-2x) - \cos(4x)) = \frac{1}{2} (\cos 2x - \cos 4x) $$ Substitute this back into the integral: $$ I = \int_0^{\pi/2} \sin 2x \cdot \frac{1}{2} (\cos 2x - \cos 4x) \,dx $$ $$ I = \frac{1}{2} \int_0^{\pi/2} (\sin 2x \cos 2x - \sin 2x \cos 4x) \,dx $$ Step 2: Simplify the integrand using further identities. For the first term, $\sin 2x \cos 2x$, use the identity $\sin 2A = 2 \sin A \cos A$, which implies $\sin A \cos A = \frac{1}{2} \sin 2A$: $$ \sin 2x \cos 2x = \frac{1}{2} \sin(2 \cdot 2x) = \frac{1}{2} \sin 4x $$ For the second term, $\sin 2x \cos 4x$, use the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$: $$ \sin 2x \cos 4x = \frac{1}{2} (\sin(2x+4x) + \sin(2x-4x)) = \frac{1}{2} (\sin 6x + \sin(-2x)) = \frac{1}{2} (\sin 6x - \sin 2x) $$ Substitute these simplified terms back into the integral expression: $$ I = \frac{1}{2} \int_0^{\pi/2} \left( \frac{1}{2} \sin 4x - \frac{1}{2} (\sin 6x - \sin 2x) \right) \,dx $$ $$ I = \frac{1}{4} \int_0^{\pi/2} (\sin 4x - \sin 6x + \sin 2x) \,dx $$ Step 3: Integrate and evaluate the definite integral. Integrate each term using the formula $\int \sin(ax) \,dx = -\frac{\cos(ax)}{a}$: $$ I = \frac{1}{4} \left[ -\frac{\cos 4x}{4} + \frac{\cos 6x}{6} - \frac{\cos 2x}{2} \right]_0^{\pi/2} $$ Now, evaluate the expression at the limits of integration: $$ I = \frac{1}{4} \left[ \left( -\frac{\cos(4 \cdot \pi/2)}{4} + \frac{\cos(6 \cdot \pi/2)}{6} - \frac{\cos(2 \cdot \pi/2)}{2} \right) - \left( -\frac{\cos(0)}{4} + \frac{\cos(0)}{6} - \frac{\cos(0)}{2} \right) \right] $$ $$ I = \frac{1}{4} \left[ \left( -\frac{\cos(2\pi)}{4} + \frac{\cos(3\pi)}{6} - \frac{\cos(\pi)}{2} \right) - \left( -\frac{1}{4} + \frac{1}{6} - \frac{1}{2} \right) \right] $$ Substitute the known values for cosine: $\cos(2\pi)=1$, $\cos(3\pi)=-1$, $\cos(\pi)=-1$, $\cos(0)=1$. $$ I = \frac{1}{4} \left[ \left( -\frac{1}{4} + \frac{-1}{6} - \frac{-1}{2} \right) - \left( -\frac{1}{4} + \frac{1}{6} - \frac{1}{2} \right) \right] $$ $$ I = \frac{1}{4} \left[ \left( -\frac{1}{4} - \frac{1}{6} + \frac{1}{2} \right) - \left( -\frac{1}{4} + \frac{1}{6} - \frac{1}{2} \right) \right] $$ Combine the fractions within each parenthesis using a common denominator of 12: $$ I = \frac{1}{4} \left[ \left( \frac{-3 - 2 + 6}{12} \right) - \left( \frac{-3 + 2 - 6}{12} \right) \right] $$ $$ I = \frac{1}{4} \left[ \frac{1}{12} - \left( -\frac{7}{12} \right) \right] $$ $$ I = \frac{1}{4} \left[ \frac{1}{12} + \frac{7}{12} \right] $$ $$ I = \frac{1}{4} \left[ \frac{8}{12} \right] $$ $$ I = \frac{1}{4} \cdot \frac{2}{3} $$ $$ I = \frac{2}{12} = \frac{1}{6} $$