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प्रश्न
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उत्तर
\[Let\ I = \int_0^\pi \frac{1}{5 + 3 \cos x} d\ x . Then, \]
\[I = \int_0^\pi \frac{1}{5 + 3\left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)} d x\]
\[ \Rightarrow I = \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{5 + 5 \tan^2 \frac{x}{2} - 3 \tan^2 \frac{x}{2}} dx\]
\[ \Rightarrow I = \int_0^\pi \frac{\sec^2 \frac{x}{2}}{5 + 2 \tan^2 \frac{x}{2}} dx\]
\[Let\ \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^2 \frac{x}{2} dx = dt\]
\[When\ x = 0, t = 0\ and\ x = \pi, t = \infty \]
\[ \therefore I = \int_0^\infty \frac{dt}{5 + 2 t^2}\]
\[ \Rightarrow I = \frac{1}{2} \int_0^\infty \frac{dt}{\frac{5}{2} + t^2}\]
\[ \Rightarrow I = \frac{1}{2} \left[ \tan^{- 1} \frac{\sqrt{5}t}{\sqrt{2}} \right]_0^\infty \]
\[ \Rightarrow I = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right)\]
\[ \Rightarrow I = \frac{\pi}{4}\]
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