The value of int sin 4x cos 3x , dx is | vedclass

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(A) Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$\sin 4x \cos 3x = \frac{1}{2} [\sin(4x+3x) + \sin(4x-3x)] = \frac{1}{2

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$-\frac{1}{14} \cos 7x - \frac{1}{2} \cos x + c$

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(A) Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$\sin 4x \cos 3x = \frac{1}{2} [\sin(4x+3x) + \sin(4x-3x)] = \frac{1}{2} [\sin 7x + \sin x]$Now, integrate the expression:$\int \sin 4x \cos 3x \, dx = \frac{1}{2} \int (\sin 7x + \sin x) \, dx$$= \frac{1}{2} [-\frac{\cos 7x}{7} - \cos x] + c$$= -\frac{1}{14} \cos 7x - \frac{1}{2} \cos x + c$

The value of $\int \sin 4x \cos 3x \, dx$ is

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