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प्रश्न
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उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{\cos^2 x + 3 \cos x + 2} d x . Then, \]
\[Let\ \cos x = t . Then, - \sin\ x\ dx\ = dt\]
\[When\ x = 0, t = 1\ and\ x\ = \frac{\pi}{2}, t = 0\]
\[ \therefore I = - \int_1^0 \frac{t dt}{t^2 + 3t + 2}\]
\[ \Rightarrow I = \int_1^0 \frac{- t dt}{\left( t + 2 \right)\left( t + 1 \right)}\]
\[ \Rightarrow I = \int_1^0 \left( \frac{1}{\left( t + 1 \right)} - \frac{2}{\left( t + 2 \right)} \right) dt\]
\[ \Rightarrow I = \left[ \log \left( t + 1 \right) - 2 \log \left( t + 2 \right) \right]_1^0 \]
\[ \Rightarrow I = \left[ \log \frac{\left( t + 1 \right)}{\left( t + 2 \right)^2} \right]_0^1 \]
\[ \Rightarrow I = \left[ \log \left( \frac{1}{4} \right) - \log \left( \frac{2}{9} \right) \right]_0^1 \]
\[ \Rightarrow I = \log \frac{9}{8}\]
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