Let (2- x)/(3+ x)dx=(1)/(2)( x+ _e| x+ x|)+C, where C is the constant of integration.

Question

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

Accepted Answer

$I=\int \frac{2-	an x}{3+	an x}dx=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$

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

$\begin{gathered} a+3b=2\, \ -3a+b=-1\, \end{gathered}$

From equation (i) and (ii) $a=b=\frac{1}{2}$

$I=\frac{1}{2}\int \frac{-3\sin x+\cos x}{3\cos x+\sin x}dx+\frac{1}{2}\int \frac{3\cos x+\sin x}{3\cos x+\sin x}dx$

$I=\frac{1}{2}\ln |3\cos x+\sin x|+\frac{1}{2}x=\frac{1}{2}(dx+\log |\beta \sin x+\gamma \cos x|)$

On comparing, we get

$I=\frac{1}{2}\ln |3\cos x+\sin x|+\frac{1}{2}x=\frac{1}{2}(dx+\log |\beta \sin x+\gamma \cos x|)$

On comparing, we get

$\begin{gathered} \Rightarrow \frac{1}{2}(x+\log (|3\cos x+\sin x|))= \ \, \frac{1}{2}(dx+\log |\beta \sin x+\gamma \cos x|) \end{gathered}$

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

Solutions · 1