Integrate the rational function in exercise: (1- )(2- ) [Hint: Pu… - Vidyadip

Question

Integrate the rational function in exercise:
$\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}$
[Hint: Put sin x = t]

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Answer

$\text{I}=\int\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}\text{dx}\dots(\text{i})$
Putting $\sin \text{x = t}$
$\Rightarrow \ \cos\text{x}=\frac{\text{dx}}{\text{dt}}$
$\Rightarrow \cos \text{x dx = dt}$
$\therefore$ From eq. (i),
$\text{I}=\int\frac{1}{(1-\text{t})(2-\text{t})}\text{dt}=\int\frac{(2-\text{t})-(1-\text{t})}{(1-\text{t})(2-\text{t})}\text{dt}$
$=\int\Bigg(\frac{(2-\text{t})}{(1-\text{t})(2-\text{t})}-\frac{(1-\text{t})}{(1-\text{t})(2-\text{t})}\Bigg)\text{dt}=\int\Bigg(\frac{1}{(1-\text{t})}-\frac{1}{(2-\text{t})}\Bigg)\text{dt}$
$=\int\frac{1}{1-\text{t}}\text{dt}-\int\frac{1}{2-\text{t}}\text{dt}=\frac{\text{log}|1-\text{t}|}{-1\rightarrow\text{Coeff. of t}}-\frac{\text{log}|2-\text{t}}{-1}+\text{c}$
$=-\text{log}|1-\text{t}|+\text{log}|2-\text{t}|+\text{c}=\text{log}\Bigg|\frac{2-\text{t}}{1-\text{t}}\Bigg|+\text{c}=\text{log}\Bigg|\frac{2-\sin\text{x}}{1-\sin\text{x}}\Bigg|+\text{c}$

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