Let [ (2- x)/(3+ x) ~d x=(1)/(2)( x+ _e| x+ x|)+C], where C is the constant of integration…

Question

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 :

Accepted Answer

$$I=\int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x$$

Put, $$2 \cos x-\sin x=a(-3 \sin x+\cos x)+ b(3 \cos x+\sin x)$$

$$\begin{aligned}
& a+3 b=2 \quad \text{.... (i)}\
& -3 a+b=-1 \quad \text{.... (ii)}
\end{aligned}$$

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

$$\begin{aligned}
& \Rightarrow \frac{1}{2}(x+\log (|3 \cos x+\sin x|))= \
& \qquad \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)
\end{aligned}$$

$$\begin{aligned}
& \alpha=1, \beta=1, \gamma=3 \
& \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4
\end{aligned}$$

Solutions · 1