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

Accepted Answer

Answer: D. 4 I=\int \frac{2-	an x}{3+	an x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x Put, 2 \cos x-\sin x=a(-3 \sin x+\cos x)+ b(3 \cos x+\sin x) \begin{aligned} & a+3 b=2 \quad 	ext{.... (i)}\ & -3 a+b=-1 \quad 	ext{.... (ii)} \end{aligned} From equation (i) and (ii) a=b=\frac{1}{2} I=\frac{1}{2} \int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\frac{1}{2} \int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x= \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|) On comparing, we get I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x=\frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|) On comparing, we get \begin{aligned} & \Rightarrow \frac{1}{2}(x+\log (|3 \cos x+\sin x|))= \ & \qquad \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|) \end{aligned} \begin{aligned} & \alpha=1, \beta=1, \gamma=3 \ & \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4 \end{aligned}

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

Answer

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