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