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प्रश्न
\[\int x \cos^2 x\ dx\]
बेरीज
उत्तर
\[\int x \cos^2 x dx\]
` " Taking x as the first function and cos"^2 x " as the second function ." `
\[ = x\int\frac{1 + \cos 2x}{2}dx - \int\left\{ \frac{d}{dx}\left( x \right)\int\frac{1 + \cos 2x}{2}dx \right\}dx\]
\[ = \frac{x}{2}\left[ x + \frac{\sin2x}{2} \right] - \int\frac{1}{2}\left( x + \frac{\sin2x}{2} \right)dx\]
\[ = \frac{x}{2}\left[ x + \frac{\sin2x}{2} \right] - \left[ \frac{x^2}{4} - \frac{\cos2x}{8} \right] + C\]
\[ = \frac{x^2}{2} + \frac{x \sin2x}{2} - \frac{x^2}{4} + \frac{\cos2x}{8} + C\]
\[ = \frac{x^2}{4} + \frac{x \sin2x}{2} + \frac{\cos2x}{8} + C\]
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