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Question Number 129223 by bramlexs22 last updated on 14/Jan/21

$$\mathrm{W}=\int \frac{\cos 5\mathrm{x}+\cos 4\mathrm{x}}{2\cos 3\mathrm{x}-1}\mathrm{dx}$$

Answered by bobhans last updated on 14/Jan/21

$$W=\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)}{2\cos 3x-1}dx$$$$W=\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right).\sin 3x}{2\cos 3x\sin 3x-\sin 3x}dx$$$$W=\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)\sin 3x}{\sin 6x-\sin 3x}dx$$$$W=\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)2\sin \left(\frac{3x}{2}\right)\cos \left(\frac{3x}{2}\right)}{-2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{3x}{2}\right)}dx$$$$W=-\int 2\sin \left(\frac{3x}{2}\right)\cos \left(\frac{x}{2}\right)dx$$$$W=-\int (\sin 2x+\sin x)dx$$$$\underset{―}{W=\frac{1}{2}\cos 2x+\cos x+C}$$

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