Question
Evaluvate the following intregals:
$\int\frac{2\tan\text{x}+3}{3\tan\text{x}+4}\ \text{dx}$
$\int\frac{2\tan\text{x}+3}{3\tan\text{x}+4}\ \text{dx}$
✓
Answer
Let $\text{I}=\int\frac{2\tan\text{x}+3}{3\tan\text{x}+4}\ \text{dx}$
$=\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
Let $2\sin\text{x}+3\cos\text{x}=\lambda\frac{\text{d}}{\text{dx}}(3\sin\text{x}+4\cos\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$2\sin\text{x}+3\cos\text{x}=\lambda(3\cos\text{x}-4\sin\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$2\sin\text{x}+3\cos\text{x}=(3\lambda+4\mu)\cos\text{x}+(-4\lambda+3\mu)\sin\text{x}+\text{v}$
Comparing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$3\lambda+4\mu=3\ \dots\dots(1)$
$-4\lambda+3\mu=2\ \dots\dots(2)$
Solving the equation (1), (2) and (3),
$\mu=\frac{18}{25}$
$\lambda=\frac{1}{25}$
$\text{V}=0$
$\text{I}=\frac{1}{25}\int\frac{(3\cos\text{x}-4\sin\text{x})}{(3\sin\text{x}+4\cos\text{x})}\ \text{dx}+\frac{18}{25}\int\text{dx}$
$\text{I}=\frac{18}{25}\text{x}+\frac{1}{25}\log|3\sin\text{x}+4\cos\text{x}|+\text{C}$
$=\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
Let $2\sin\text{x}+3\cos\text{x}=\lambda\frac{\text{d}}{\text{dx}}(3\sin\text{x}+4\cos\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$2\sin\text{x}+3\cos\text{x}=\lambda(3\cos\text{x}-4\sin\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$2\sin\text{x}+3\cos\text{x}=(3\lambda+4\mu)\cos\text{x}+(-4\lambda+3\mu)\sin\text{x}+\text{v}$
Comparing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$3\lambda+4\mu=3\ \dots\dots(1)$
$-4\lambda+3\mu=2\ \dots\dots(2)$
Solving the equation (1), (2) and (3),
$\mu=\frac{18}{25}$
$\lambda=\frac{1}{25}$
$\text{V}=0$
$\text{I}=\frac{1}{25}\int\frac{(3\cos\text{x}-4\sin\text{x})}{(3\sin\text{x}+4\cos\text{x})}\ \text{dx}+\frac{18}{25}\int\text{dx}$
$\text{I}=\frac{18}{25}\text{x}+\frac{1}{25}\log|3\sin\text{x}+4\cos\text{x}|+\text{C}$
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