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प्रश्न
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
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उत्तर
\[\int_0^\frac{\pi}{2} \frac{\cos x}{1 + \sin^2 x} d x\]
\[Let \sin x = t,\text{ then }\cos x dx = dt\]
\[\text{When }x \to 0 ; t \to 0\]
\[\text{And }x \to \frac{\pi}{2}; t \to 1\]
Therefore the integral becomes
\[ \int_0^1 \frac{dt}{1 + t^2}\]
\[ = \left[ \tan^{- 1} x \right]_0^1 \]
\[ = \frac{\pi}{4}\]
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