Question

\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]

Answer 1

\[\text{ Let  I} = \int\left( 1 + x^2 \right) \cdot \cos 2x \cdot dx\]
\[ = \int\text{ cos  2x  dx } + \int x^2 \cdot \text{ cos  2x  dx }\]
\[ = \frac{\sin 2x}{2} + I_1 \text{ where I}_1 = \int x^2\text{ cos  2x  dx } . . . \left( 1 \right)\]
\[ I_1 = \int {x^2}_I \cos_{II} 2x \text{  dx }\]
\[ = x^2 \int\text{ cos  2x  dx } - \int\left\{ \frac{d}{dx}\left( x^2 \right)\int \text{ cos  2x  dx } \right\}dx\]
\[ = x^2 \cdot \frac{\sin 2x}{2} - \int\frac{2x \times \sin 2x}{2} dx\]
\[ = \frac{x^2 \cdot \sin 2x}{2} - \int x_I \cdot \sin_{II} 2x\text{  dx }\]
\[ = \frac{x^2 \cdot \sin 2x}{2} - \left[ x\int \text{ sin 2x dx }- \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin 2x dx} \right\}dx \right]\]
\[ = \frac{x^2 \cdot \sin 2x}{2} - \left[ x\left( \frac{- \cos 2x}{2} \right) - \int1 \cdot \left( \frac{- \cos 2x}{2} \right) dx \right]\]
\[ = \frac{x^2 \cdot \sin 2x}{2} + \frac{x \cdot \cos 2x}{2} - \frac{\sin 2x}{4} . . . \left( 2 \right)\]
\[\text{ From }\left( 1 \right) \text{ and }\left( 2 \right)\]
\[ \therefore I = \frac{\sin 2x}{2} + \frac{x^2 \sin 2x}{2} + \frac{x \cos 2x}{2} - \frac{\sin 2x}{4} + C\]
\[ = \left( x^2 + 1 \right) \frac{\sin 2x}{2} + \frac{x \cos 2x}{2} - \frac{\sin 2x}{4} + C\]