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3.2(A) · Q52

Q.Integrate: ∫cos⁡3x−cos⁡4xsin⁡3x+sin⁡4x dx\int \dfrac{\cos 3x-\cos 4x}{\sin 3x+\sin 4x}\,dx

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Sum-to-product: cos⁡3x−cos⁡4x=2sin⁡7x2sin⁡x2\cos3x-\cos4x=2\sin\dfrac{7x}2\sin\dfrac x2, and sin⁡3x+sin⁡4x=2sin⁡7x2cos⁡x2\sin3x+\sin4x=2\sin\dfrac{7x}2\cos\dfrac x2.

The ratio is 2sin⁡7x2sin⁡x22sin⁡7x2cos⁡x2=tan⁡x2\dfrac{2\sin\frac{7x}2\sin\frac x2}{2\sin\frac{7x}2\cos\frac x2}=\tan\dfrac x2 (the sin⁡7x2\sin\frac{7x}2 factor cancels). …

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