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Question Number 23796 by tapan das last updated on 06/Nov/17

$$\int \frac{2\mathrm{sinx}+3\mathrm{cosx}}{3\mathrm{sinx}+4\mathrm{cosx}}\mathrm{dx}$$

Answered by ajfour last updated on 06/Nov/17

$$2\sin x+3\cos x=A(3\sin x+4\cos x)$$$$+B(3\cos x-4\sin x)$$$$\Rightarrow 3A-4B=2and$$$$4A+3B=3$$$$\Rightarrow A=\frac{18}{25},B=\frac{1}{25};So$$$$\int \frac{2\sin x+3\cos x}{3\sin x+4\cos x}dx=\frac{18}{25}\int dx+$$$$+\frac{1}{25}\int \frac{3\cos x-4\sin x}{3\sin x+4\cos x}dx$$$$=\frac{18x}{25}+\frac{1}{25}\ln \mid 3\sin x+4\cos x\mid +C.$$

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