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प्रश्न
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उत्तर
\[Let\ I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \sin^2 x d x\]
\[Here\ f\left( x \right) = \sin^2 x\]
\[f\left( - x \right) = \sin^2 \left( - x \right) = \sin^2 x = f\left( x \right)\]
\[\text{Hence} \sin^2 x \text{is an even function}\]
Therefore,
\[I = 2 \int_0^\frac{\pi}{4} \sin^2 x d x\]
\[ = 2 \int_0^\frac{\pi}{4} \left( \frac{1 - \cos2x}{2} \right)dx\]
\[ = \int_0^\frac{\pi}{4} \left( 1 - \cos2x \right) dx\]
\[ = \left[ x - \frac{\sin2x}{2} \right]_0^\frac{\pi}{4} \]
\[ = \frac{\pi}{4} - \frac{1}{2}\]
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