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