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Integrate the rational functions
$\frac{\cos x}{(1-\sin x)(2-\sin x)}$
$\frac{\cos x}{(1-\sin x)(2-\sin x)}$
Solution:
1760 Upvotes
Verified Answer
Put $\sin x=t$, so that $\cos x d x=d t$
$\therefore \mathrm{I}=\int \frac{1}{(1-t)(2-t)} d t \quad \ldots(i)$
Let $\frac{1}{(1-t)(2-t)} \equiv \frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{2-t}$
$\Rightarrow 1 \equiv \mathrm{A}(2-t)+\mathrm{B}(1-t) \quad \ldots(ii)$
Put, $t=1,2$ in (ii), we get : $\mathrm{A}=1 \& \mathrm{~B}=-1$
$\begin{aligned}
&\therefore \mathrm{I}=\int \frac{1}{1-t} d t-\int \frac{d t}{2-t} \\
&=\log \left|\frac{2-t}{1-t}\right|+\mathrm{C}=\log \left|\frac{2-\sin x}{1-\sin x}\right|+\mathrm{C}
\end{aligned}$
$\therefore \mathrm{I}=\int \frac{1}{(1-t)(2-t)} d t \quad \ldots(i)$
Let $\frac{1}{(1-t)(2-t)} \equiv \frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{2-t}$
$\Rightarrow 1 \equiv \mathrm{A}(2-t)+\mathrm{B}(1-t) \quad \ldots(ii)$
Put, $t=1,2$ in (ii), we get : $\mathrm{A}=1 \& \mathrm{~B}=-1$
$\begin{aligned}
&\therefore \mathrm{I}=\int \frac{1}{1-t} d t-\int \frac{d t}{2-t} \\
&=\log \left|\frac{2-t}{1-t}\right|+\mathrm{C}=\log \left|\frac{2-\sin x}{1-\sin x}\right|+\mathrm{C}
\end{aligned}$
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