Question

int(1 + x2) cos 2x dx​

Answer 1

Step-by-step explanation:

We have,

$\int(1 + {x}^{2} ) \cos(2x) dx$

$= \int \cos(2x) dx + \int {x}^{2} \cos(2x) dx$

$= \frac{ \sin(2x) }{2} + {x}^{2} \int \cos(2x) dx - \int( \frac{d}{dx} {(x)}^{2} .\int \cos(2x) dx)dx$

$= \sin(x) \cos(x) + \frac{ {x}^{2} \sin(2x) }{2} - \int2x. \frac{ \sin(2x) }{2} dx$

$= \sin(x) \cos(x) + {x}^{2} \sin(x) \cos(x) - (x\int \sin(2x) dx - \int( \frac{d}{dx}(x) .\int \sin(2x) dx)dx)$

$= (1 + {x}^{2} ) \sin(x) \cos(x) - ( - \frac{x \cos(2x) }{2} + \int \frac{ \cos(2x) }{2} )$

$= (1 + {x}^{2} ) \sin(x) \cos(x) + \frac{x \cos(2x) }{2} - \frac{ \sin(2x) }{2} + c$

$= {x}^{2}\sin(x) \cos(x) + \frac{x \cos(2x) }{2} + c$