If int sin 5x cos 3x ; dx = - frac{cos 8x}{16 | vedclass

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(B) We use the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$.$\int \sin 5x \cos 3x \; dx = \frac{1}{2} \int (\sin(5x+3x) + \sin(5x-

Vedclass · Accepted Answer

$-\frac{\cos 2x}{4} + 	ext{constant}$

Vedclass · Answer

(B) We use the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$.$\int \sin 5x \cos 3x \; dx = \frac{1}{2} \int (\sin(5x+3x) + \sin(5x-3x)) \; dx$$= \frac{1}{2} \int (\sin 8x + \sin 2x) \; dx$$= \frac{1}{2} \left( -\frac{\cos 8x}{8} - \frac{\cos 2x}{2} \right) + C$$= -\frac{\cos 8x}{16} - \frac{\cos 2x}{4} + C$Comparing this with the given expression $-\frac{\cos 8x}{16} + A$,we find that $A = -\frac{\cos 2x}{4} + C$.

If $\int \sin 5x \cos 3x \; dx = - \frac{\cos 8x}{16} + A$,then $A = $

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