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Π / 2 ∫ 0 X 2 Cos 2 X D X - Mathematics

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प्रश्न

\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]

बेरीज
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उत्तर

\[\int_0^\frac{\pi}{2} x^2 \cos2x d x\]

\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2x \frac{\sin2x}{2}dx\]

\[ = \left[ x^2 \frac{sin2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} x \sin 2x dx\]

\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} - \left[ - x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} + \left[ - \int_0^\frac{\pi}{2} \frac{\cos2x}{2}dx \right]\]

\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} + \left[ x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} \frac{\cos2x}{2}dx\]

\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} + \left[ x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} - \frac{1}{2} \left[ \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]

\[ = 0 - \frac{\pi}{4} - 0\]

\[ = \frac{- \pi}{4}\]

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Definite Integrals
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पाठ 20: Definite Integrals - Revision Exercise [पृष्ठ १२१]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Revision Exercise | Q 21 | पृष्ठ १२१

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