Let $\int \frac{2- an x}{3+ an x} \mathrm{~d} x=\frac{1}{2}\left(\alpha x+\log _{\mathrm{e}}|\beta \sin x+\gamma \cos x| ight)+C$, where $C$ is the constant of integration. Then $\alpha+\frac{\gamma}{\beta}$ is equal to :

Question

Let $\int \frac{2-	an x}{3+	an x} \mathrm{~d} x=\frac{1}{2}\left(\alpha x+\log _{\mathrm{e}}|\beta \sin x+\gamma \cos x|ight)+C$, where $C$ is the constant of integration. Then $\alpha+\frac{\gamma}{\beta}$ is equal to :

Aicharya · Accepted Answer

$\begin{aligned} & \int \frac{2-	an x}{3+	an x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x \ & 2 \cos x-\sin x=A(3 \cos x+\sin x)+B(\cos x-3 \sin x) \ & 3 A+B=2 \ & A-3 B=-1\end{aligned}$ $\begin{aligned} & \Rightarrow A=\frac{1}{2}, B=\frac{1}{2} \ & 	herefore \int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x \ & =\frac{x}{2}+\frac{1}{2} \ln |3 \cos x+\sin x|+C \ & =\frac{1}{2}(x+\ln |3 \cos x+\sin x|)+C \ & =\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C \ & \alpha=1, \beta=1, \gamma=3 \ & 	herefore \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4\end{aligned}$

Let $\int \frac{2-\tan x}{3+\tan x} \mathrm{~d} x=\frac{1}{2}\left(\alpha x+\log _{\mathrm{e}}|\beta \sin…

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