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Question Number 82604 by jagoll last updated on 23/Feb/20

$$\int \frac{\cos \left(5x\right)+\cos \left(4x\right)dx}{1-2\cos \left(3x\right)}=$$

Answered by john santu last updated on 23/Feb/20

$$\int \frac{\{\cos \left(5x\right)+\cos \left(4x\right)\}\sin \left(3x\right)}{\sin \left(3x\right)-\sin \left(6x\right)}dx$$$$=2\int \frac{\left\{\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)\right\}\sin \left(3x\right)}{2\cos \left(\frac{9x}{2}\right)\sin (-\frac{3x}{2})}dx$$$$=-\int \frac{\sin \left(3x\right)\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{3x}{2}\right)}dx$$$$=-\int \frac{2\sin \left(\frac{3x}{2}\right)\cos \left(\frac{3x}{2}\right)\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{3x}{2}\right)}dx$$$$=-\int 2\cos \left(\frac{3x}{2}\right)\cos \left(\frac{x}{2}\right)dx$$$$=-\int \{\cos 2x+\cos x\}dx$$$$=-\{\frac{1}{2}\sin 2x-\sin x\}+c$$$$=-\sin x\cos x+\sin x+c$$$$=\sin x(1-\cos x)+c$$

Commented by jagoll last updated on 23/Feb/20

$$thankyousir$$

Commented by peter frank last updated on 23/Feb/20

$$thankyou$$

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