Question

$\mathrm{If}\int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx=ax+b\ln \left(2\sin x+3\cos x\right|+C,\mathrm{then}$

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• A. $a=-\frac{12}{13},b=\frac{15}{39}$
• B. $a=-\frac{7}{13},b=\frac{6}{39}$
• C. $a=\frac{12}{13},b=-\frac{15}{39}$

Answer 1

The correct option is C $a=\frac{12}{13},b=-\frac{15}{39}$

Here, in order to find the values of $a\mathrm{and}b$
We will follow the following approach$\int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx=ax+b\ln \left(2\sin x+3\cos x\right|+C\mathrm{Differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\frac{d}{dx}\left(ax+b\ln \left(2\sin x+3\cos x\right|\right)=\frac{3\sin x+2\cos x}{3\cos x+2\sin x}\Rightarrow a+b\frac{2\cos x-3\sin x}{2\sin x+3\cos x}=\frac{3\sin x+2\cos x}{3\cos x+2\sin x}\Rightarrow \frac{(2a-3b)\sin x+(3a+2b)\cos x}{2\sin x+3\cos x}=\frac{3\sin x+2\cos x}{3\cos x+2\sin x}\mathrm{Comparing}\mathrm{we}\mathrm{get},2a-3b=3\mathrm{and}3a+2b=2\mathrm{Solving}\mathrm{these}\mathrm{two}\mathrm{equation}\mathrm{we}\mathrm{get},a=\frac{12}{13}\mathrm{and}b=\frac{-15}{39}$