Question
Integrate the functions in Exercises:
$\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}$
$\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}$
✓
Answer
$\text{Let I }=\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{ dx}=\int\frac{2\cos\text{x}-3\sin\text{x}}{2(2\sin\text{x}+3\cos\text{x})}\text{ dx} $
$=\frac{1}{2}\int\frac{2\cos\text{x}-3\sin\text{x}}{2\sin\text{x}+3\cos\text{x}}\text{ dx} \ \ \ \ \ \ ...\text{(i)} $
Putting $2\sin\text{x}+3\cos\text{x}=\text{t}\ \ \ \ \Rightarrow \ \ \ \ 2\cos\text{x}-3\sin\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \ \ \ \ (2\cos\text{x}-3\sin\text{x})\text{ dx}=\text{dt} $
$ \therefore \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}} =\frac{1}{2}\log\begin{vmatrix}t\end{vmatrix}+\text{c}$
$=\frac{1}{2}\log\begin{vmatrix}2\sin\text{x}+3\cos \text{x}\end{vmatrix}+\text{c}$
$=\frac{1}{2}\int\frac{2\cos\text{x}-3\sin\text{x}}{2\sin\text{x}+3\cos\text{x}}\text{ dx} \ \ \ \ \ \ ...\text{(i)} $
Putting $2\sin\text{x}+3\cos\text{x}=\text{t}\ \ \ \ \Rightarrow \ \ \ \ 2\cos\text{x}-3\sin\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \ \ \ \ (2\cos\text{x}-3\sin\text{x})\text{ dx}=\text{dt} $
$ \therefore \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}} =\frac{1}{2}\log\begin{vmatrix}t\end{vmatrix}+\text{c}$
$=\frac{1}{2}\log\begin{vmatrix}2\sin\text{x}+3\cos \text{x}\end{vmatrix}+\text{c}$
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