The value of the integral ∫sin 3x cos 4x dx is

Let I=∫sin 3xcos 4xdx=12∫2sin 3xcos 4xdx=12∫{sin (3x+4x)+sin (3x−4x)}dx=12∫(sin 7x-sinx) dx (∵2sin Acos B=sin (A+B)+sin (A−B))=12−cos 7x7+cos x+C ∵∫sin aθdx=−cos aθa=−cos 7x14+cos x2+C

The value of the integral ∫sin 3x cos 4x dx is

Let I=∫sin 3xcos 4xdx=12∫2sin 3xcos 4xdx=12∫{sin (3x+4x)+sin (3x−4x)}dx=12∫(sin 7x-sinx) dx (∵2sin Acos B=sin (A+B)+sin (A−B))=12−cos 7x7+cos x+C ∵∫sin aθdx=−cos aθa=−cos 7x14+cos x2+C

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

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

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

$\frac{\cos 7 x}{14} + \frac{\cos x}{2} + C$

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

answer is D.

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Detailed Solution

Let $\begin{array}{l} I = \int \sin 3 xcos 4 xdx = \frac{1}{2} \int 2 \sin 3 xcos 4 xdx \\ = \frac{1}{2} \int {\sin (3 x + 4 x) + \sin (3 x - 4 x)} dx \\ = \frac{1}{2} \int (\sin 7 x - \sin x) d x (∵ 2 \sin Acos B = \sin (A + B) + \sin (A - B)) \\ = \frac{1}{2} (\frac{- \cos 7 x}{7} + \cos x) + C [∵ \int \sin aθdx = \frac{- \cos aθ}{a}] \\ = \frac{- \cos 7 x}{14} + \frac{\cos x}{2} + C \end{array}$