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Question Number 163072 by abdullahhhhh last updated on 03/Jan/22

$$\int \frac{\mathbf{sinx}+\mathbf{sin}3\mathbf{x}+\mathbf{sin}5\mathbf{x}+\mathbf{sin}7\mathbf{x}}{\mathbf{cosx}+\mathbf{cos}3\mathbf{x}+\mathbf{cos}5\mathbf{x}+\mathbf{cos}7\mathbf{x}}\mathbf{dx}$$

Answered by tounghoungko last updated on 03/Jan/22

$$\frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}$$$$=\frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}$$$$=\frac{\sin 4x}{\cos 4x}$$$$\int \frac{d(-\frac{1}{4}\cos 4x)}{\cos 4x}=-\frac{1}{4}\ln \mid \cos 4x\mid +c$$

Commented by tounghoungko last updated on 03/Jan/22

$$\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$$

Commented by Ar Brandon last updated on 03/Jan/22

$$\frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}$$$$=\frac{2\sin 4x\cancel{(\cos 3x+\cos x)}}{2\cos 4x\cancel{(\cos 3x+\cos x)}}=\frac{\sin 4x}{\cos 4x}$$

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