Question
Evaluvate the following intregals:
$\int\frac{8\cot\text{x}+1}{3\cot\text{x}+2}\ \text{dx}$
$\int\frac{8\cot\text{x}+1}{3\cot\text{x}+2}\ \text{dx}$
$=\int\bigg(\frac{8\frac{\cos\text{x}}{\sin\text{x}}+1}{\frac{3\cos\text{x}}{\sin\text{x}}+2}\bigg)\text{dx}$
$=\int\Big(\frac{8\cos\text{x}+\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
Now, Let $8\cos\text{x}+\sin\text{x}=\text{A}(3\cos\text{x}+2\sin\text{x})+\text{B}(-3\sin\text{x}+2\cos\text{x})\ \dots(1)$
$\Rightarrow8\cos\text{x}+\sin\text{x}+\sin\text{x}=3\text{A}\cos\text{x}+2\text{A}\sin\text{x}-3\text{B}\sin\text{x}+2\text{B}\cos\text{x}$
$\Rightarrow8\cos\text{x}+\sin\text{x}-(3\text{A}+2\text{B})\cos\text{x}+(2\text{A}-3\text{B})\sin\text{x}$
Equating the coefficient of like terms we get,
$2\text{A}-3\text{B}=1\ \dots(2)$
$3\text{A}+2\text{B}=8\ \dots(3)$
Solving eq (2) and eq (3) we get,
$\text{A}=2,\text{B}=1$
Thus, by substracting the value of A and B in eq (1) we get,
$\text{I}=\int\Big[\frac{2(3\cos\text{x}+2\sin\text{x})+1(-3\sin\text{x}+2\cos\text{x})}{(3\cos\text{x}+2\sin\text{x})}\Big]\text{dx}$
$=2\int\Big(\frac{3\cos\text{x}+2\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}+\int\Big(\frac{-3\sin\text{x}+2\cos\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
$=2\int\text{dx}+\int\Big(\frac{-3\sin\text{x}+2\cos\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
Putting
$3\cos\text{x}+2\sin\text{x}=\text{t}$$\Rightarrow(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=2\int\text{dx}+\int\frac{1}{\text{t}}\text{dt}$
$=2\text{x}+\ln|\text{t}|+\text{C}$
$=2\text{x}+\ln|3\cos\text{x}+2\sin\text{x}|+\text{C}$
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| Differential equation | Function |
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