Question

$\int xcos^{2} x dx$

Answer 1

[tex]\it I = \int x \cos^2x\ dx = \int x \dfrac{1+\cos2x}{2}\ dx = \dfrac{1}{2}\int x(1+cos2x)\ dx = \\\;\\ \\\;\\ = \dfrac{1}{2} (\int x\ dx + \int x cos2x\ dx ) = \dfrac{1}{2} ( I_1 + I_2 ) [/tex]

$\it I_1 = \int x\ dx = \dfrac{1}{2} x^2 + C$

$\it I_2 = \int x \cos2x\ dx$

Folosim integrarea prin părți :

[tex]\it \int fg'=fg-\int f'g \\\;\\ f = x \Rightarrow f'=1 [/tex]

$\it g' = \cos 2x \Rightarrow g = \int cos 2x\ dx = \dfrac{1}{2} sin 2x$

[tex]\it I_2 = \dfrac{1}{2} x\sin 2x -\dfrac{1}{2} \int sin 2x\ dx = \dfrac{1}{2} x sin 2x -\dfrac{1}{2} (-\dfrac{1}{2})cos 2x + C = \\\;\\ \\\;\\ = \dfrac{1}{2} x \sin 2x + \dfrac{1}{4} cos 2x + C[/tex]

[tex]\it I = \dfrac{1}{2} ( I_1 + I_2) = \dfrac{1}{2}\left( \dfrac{1}{2} x^2 + \dfrac{1}{2}x sin 2x + \dfrac{1}{4} cos 2x \right) + C = \\\;\\ \\\;\\ = \dfrac{1}{8} (2x^2 +2x \sin2x + \cos 2x) + C[/tex]