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