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प्रश्न
Evaluate the following definite integrals:
\[\int_0^\frac{\pi}{2} x^2 \sin\ x\ dx\]
योग
उत्तर
\[\text{Let }I = \int_0^\frac{\pi}{2} x^2 \sin x\ dx\]
Applying integration by parts, we have
\[I = x^2 \left( - \cos x \right) |_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2x\left( - \cos x \right)dx\]
\[ \Rightarrow I = \left( 0 - 0 \right) + 2 \int_0^\frac{\pi}{2} x\cos x\ dx ..............\left( \cos\frac{\pi}{2} = 0 \right)\]
\[ \Rightarrow I = \left( 0 - 0 \right) + 2 \int_0^\frac{\pi}{2} x\cos x\ dx ..............\left( \cos\frac{\pi}{2} = 0 \right)\]
Again applying integration by parts, we have
\[I = 0 + 2\left[ x\sin x |_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 1 \times \sin\ x\ dx \right]\]
\[ \Rightarrow I = 2\left( \frac{\pi}{2}\sin\frac{\pi}{2} - 0 \right) - 2 \int_0^\frac{\pi}{2} \sin\ x\ dx\]
\[ \Rightarrow I = 2\left( \frac{\pi}{2} - 0 \right) - 2\left( - \cos\ x \right) |_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \pi + 2\left( \cos\frac{\pi}{2} - \cos0 \right)\]
\[ \Rightarrow I = \pi + 2\left( 0 - 1 \right)\]
\[ \Rightarrow I = \pi - 2\]
\[ \Rightarrow I = 2\left( \frac{\pi}{2}\sin\frac{\pi}{2} - 0 \right) - 2 \int_0^\frac{\pi}{2} \sin\ x\ dx\]
\[ \Rightarrow I = 2\left( \frac{\pi}{2} - 0 \right) - 2\left( - \cos\ x \right) |_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \pi + 2\left( \cos\frac{\pi}{2} - \cos0 \right)\]
\[ \Rightarrow I = \pi + 2\left( 0 - 1 \right)\]
\[ \Rightarrow I = \pi - 2\]
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