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प्रश्न
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{1 + \sin^4 x} d x . \]
\[Let\ \sin x\ = t\ . Then\, \cos x\ dx\ = dt\]
\[When\ x = 0, t = 0\ and\ x\ = \frac{\pi}{2}, t = 1\]
\[ \therefore I = \int_0^1 \frac{t}{1 + t^4} d t\]
\[Let\ t^2 = u . Then, 2t\ dt = du\]
\[So, I = \int_0^1 \frac{t}{1 + t^4} d t \]
\[ \Rightarrow I = \frac{1}{2} \int_0^1 \frac{1}{1 + u^2} d u\]
\[ \Rightarrow I = \frac{1}{2} \left[ \tan^{- 1} u \right]_0^1 \]
\[ \Rightarrow I = \frac{\pi}{8}\]
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