Question
$\int \frac{ d x}{2+3 \tan x}$
✓
Answer
$ \text { Let } I =\int \frac{1}{2+3 \tan x} d x$
$=\int \frac{1}{2+3\left(\frac{\sin x}{\cos x}\right)} d x$
$=\int \frac{\cos x}{2 \cos x+3 \sin x} d x $
Let $\cos x = A (2 \cos x+3 \sin x)+ B \frac{ d }{ d x}(2 \cos x+3 \sin x)$
$ =A(2 \cos x+3 \sin x)+B(-2 \sin x+3 \cos x)$
$\therefore \cos x+0 \cdot \sin x=\cos x(2 A+3 B)+\sin x(3 A-2 B) $
By equating the coefficients on both sides, we get
$2 A +3 B =1 \text { and } 3 A -2 B =0$
Solving these equations, we get
$ A =\frac{2}{13} \text { and } B =\frac{3}{13}$
$\therefore \cos x =\frac{2}{13}(2 \cos x+3 \sin x)+\frac{3}{13}(-2 \sin x+3 \cos x)$
$\therefore I =\int \frac{\frac{2}{13}(2 \cos x+3 \sin x)+\frac{3}{13}(-2 \sin x+3 \cos x)}{2 \cos x+3 \sin x} d x$
$\therefore I =\frac{2}{13} \int d x+\frac{3}{13} \int \frac{-2 \sin x+3 \cos x}{2 \cos x+3 \sin x} d x$
$\therefore I =\frac{2}{13} x+\frac{3}{13} \log |2 \cos +3 \sin x|+ c \ldots \ldots \ldots\left[\because \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f (x)|+ c \right] $
$=\int \frac{1}{2+3\left(\frac{\sin x}{\cos x}\right)} d x$
$=\int \frac{\cos x}{2 \cos x+3 \sin x} d x $
Let $\cos x = A (2 \cos x+3 \sin x)+ B \frac{ d }{ d x}(2 \cos x+3 \sin x)$
$ =A(2 \cos x+3 \sin x)+B(-2 \sin x+3 \cos x)$
$\therefore \cos x+0 \cdot \sin x=\cos x(2 A+3 B)+\sin x(3 A-2 B) $
By equating the coefficients on both sides, we get
$2 A +3 B =1 \text { and } 3 A -2 B =0$
Solving these equations, we get
$ A =\frac{2}{13} \text { and } B =\frac{3}{13}$
$\therefore \cos x =\frac{2}{13}(2 \cos x+3 \sin x)+\frac{3}{13}(-2 \sin x+3 \cos x)$
$\therefore I =\int \frac{\frac{2}{13}(2 \cos x+3 \sin x)+\frac{3}{13}(-2 \sin x+3 \cos x)}{2 \cos x+3 \sin x} d x$
$\therefore I =\frac{2}{13} \int d x+\frac{3}{13} \int \frac{-2 \sin x+3 \cos x}{2 \cos x+3 \sin x} d x$
$\therefore I =\frac{2}{13} x+\frac{3}{13} \log |2 \cos +3 \sin x|+ c \ldots \ldots \ldots\left[\because \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f (x)|+ c \right] $
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