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The integral is $\int\frac{\sin3x}{\cos7x\cos4x}$

I have tried sum to product rule and $\sin3x$ formula: $\int\frac{3\sin x-4\sin^3 x}{\frac{1}{2}(\cos3x+\cos11x)}$

How should I proceed further?

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$$\frac{\sin 3x}{\cos 7x\cos 4x} = \frac{\sin (7x-4x)}{\cos 7x\cos 4x}=\frac{\sin 7x\cos 4x - \cos 7x \sin 4x}{\cos 7x \cos 4x} = \tan 7x - \tan 4x$$

I think you can take it from here.

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