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प्रश्न
उत्तर
\[Let\ I = \int_0^1 \frac{\sqrt{\tan^{- 1} x}}{1 + x^2} d x . \]
\[Let\ \tan^{- 1} x = t . Then\, \frac{1}{1 + x^2} dx = dt\]
\[When\ x = 0, t = 0\ and\ x\ = 1\, t = \frac{\pi}{4}\]
\[ \therefore I = \int_0^1 \frac{\sqrt{\tan^{- 1} x}}{1 + x^2} d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{4} \sqrt{t} dt\]
\[ \Rightarrow I = \left[ \frac{2 t^\frac{3}{2}}{3} \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow I = \frac{2}{3} \left( \frac{\pi}{4} \right)^\frac{3}{2} \]
\[ \Rightarrow I = \frac{1}{12} \pi^\frac{3}{2} \]
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