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प्रश्न
पर्याय
π/4
π/2
π
1
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उत्तर
1
\[\text{We have}, \]
\[ I = \int_0^\frac{\pi}{2} x \sin x\ d x \]
\[ = \left[ - x \cos x \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 1\left( - \cos x \right) d x\]
\[ = \left[ - x \cos x \right]_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} \cos x\ d x\]
\[ = - \left[ x \cos x \right]_0^\frac{\pi}{2} + \left[ \sin x \right]_0^\frac{\pi}{2} \]
\[ = - \left[ 0 - 0 \right] + \left[ 1 - 0 \right]\]
\[ = 1\]
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