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