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प्रश्न
\[\int x \sin^3 x\ dx\]
बेरीज
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उत्तर
Let I =\[\int x \text{ sin}^3 \text{ x dx }\]
sin (3A) = 3 sin A – 4 sin3 A
\[\sin^3 A = \frac{1}{4}\left[ 3 \sin A - \sin 3A \right]\]
\[ \therefore I = \frac{1}{4}\int x . \left( 3 \sin x - \sin 3x \right)dx\]
\[ = \frac{3}{4}\int x_I . \sin_{II} \text{ x dx} - \frac{1}{4}\int x_I {. \sin_{II} \left( 3x \right)} \text{ dx }\]
\[ = \frac{3}{4}\left[ x\left( - \cos x \right) - \int1 . \left( - \cos x \right)dx \right] - \frac{1}{4}\left[ x\left( - \frac{\cos 3x}{3} \right) - \int1 . \left( - \frac{\cos 3x}{3} \right)dx \right]\]
\[ = - \frac{3x \cos x}{4} + \frac{3}{4}\sin x + \frac{x \cos 3x}{12} - \frac{1}{36}\sin 3x + C\]
\[ \therefore I = \frac{1}{4}\int x . \left( 3 \sin x - \sin 3x \right)dx\]
\[ = \frac{3}{4}\int x_I . \sin_{II} \text{ x dx} - \frac{1}{4}\int x_I {. \sin_{II} \left( 3x \right)} \text{ dx }\]
\[ = \frac{3}{4}\left[ x\left( - \cos x \right) - \int1 . \left( - \cos x \right)dx \right] - \frac{1}{4}\left[ x\left( - \frac{\cos 3x}{3} \right) - \int1 . \left( - \frac{\cos 3x}{3} \right)dx \right]\]
\[ = - \frac{3x \cos x}{4} + \frac{3}{4}\sin x + \frac{x \cos 3x}{12} - \frac{1}{36}\sin 3x + C\]
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