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प्रश्न
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
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उत्तर
\[\text { Let } \sin x = t \Rightarrow \cos x dx = dt\]
\[\int\frac{2dt}{\left( 1 - t \right)\left( 1 + t^2 \right)}\]
\[\text { Using partial fraction }\]
\[\frac{2}{\left( 1 - t \right)\left( 1 + t^2 \right)} = \frac{A}{\left( 1 - t \right)} + \frac{Bt + C}{\left( 1 + t^2 \right)}\]
\[\text { On solving } A = 1, B = 1, C = 1\]
\[\int\frac{2dt}{\left( 1 - t \right)\left( 1 + t^2 \right)} = \int\frac{dt}{\left( 1 - t \right)} + \int\frac{\left( 1 + t \right)}{\left( 1 + t^2 \right)}dt\]
\[ = \int\frac{dt}{\left( 1 - t \right)} + \int\frac{dt}{\left( 1 + t^2 \right)} + \int\frac{t dt}{\left( 1 + t^2 \right)}\]
\[ = - \ln (1 - t) + \tan^{- 1} t + \frac{1}{2}\ln\left( 1 + t^2 \right)\]
\[ = \ln \frac{\sqrt{1 + t^2}}{1 - t} + \tan^{- 1} t + C\]
\[\text { Replacing the value of t }\]
\[ = \ln \frac{\sqrt{1 + \sin^2 x}}{1 - \sin x} + \tan^{- 1} (\sin x) + C\]
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