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∫ X 2 Cos 2 X D X

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Question

\[\int x^2 \cos 2x\ \text{ dx }\]
Sum
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Solution

\[\int x^2 \text{ cos 2x dx }\]
`  " 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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Chapter 18: Indefinite Integrals - Exercise 19.25 [Page 133]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 18 Indefinite Integrals
Exercise 19.25 | Q 8 | Page 133
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