Question

Cat face integrala din xcosx totul la patrat?

Answer 1

$cos2x = cos^2x-sin^2x \Rightarrow cos2x = cos^2x-(1-cos^2x) \Rightarrow \\ \Rightarrow cos2x = cos^2x-1+cos^2x \Rightarrow cos2x = 2cos^2x-1 \Rightarrow \\ \Rightarrow 2cos^2x = cos2x+1 \Rightarrow \boxed{cos^2x = \dfrac{cos2x+1}{2} } \\ \rightarrow ne folosim de aceasta formula$

$\int\limits {(xcosx)^2} \, dx = \int\limits x^2\cdot cos^2x\, dx =\int\limits x^2\cdot \dfrac{cos2x+1}{2}\, dx =\\= \dfrac{1}{2}\int\limits x^2\cdot (cos2x+1)\, dx = \dfrac{1}{2}\int\limits (x^2\cdot cos2x+x^2)\, dx = \\ \\=\dfrac{1}{2}\int\limits x^2 \cdot cos2x\, dx + \dfrac{1}{2} \int\limits x^2\, dx = \dfrac{1}{2}\cdot \dfrac{1}{2}\int\limits 2\cdot cos2x\cdot x^2\, dx +\dfrac{1}{2}\cdot \dfrac{x^3}{3} =$
$=\dfrac{1}{4}\int\limits (sin2x)'\cdot x^2\, dx +\dfrac{x^3}{6} = \dfrac{1}{4}\cdot sin2x\cdot x^2 - \dfrac{1}{4}\int\limits sin2x\cdot 2x\, dx+\dfrac{x^3}{6} = \\ \\ =\dfrac{1}{4}\cdot sin2x\cdot x^2-\dfrac{1}{2}\int\limits sin2x\cdot x\, dx +\dfrac{x^3}{6} = \\ \\ =\dfrac{1}{4}\cdot sin2x\cdot x^2-\dfrac{1}{2}\cdot \Big(-\dfrac{1}{2}\Big)\cdot \int\limits -2\cdot sin2x\cdot x\, dx +\dfrac{x^3}{6} =$
$=\dfrac{1}{4}\cdot sin2x\cdot x^2-\dfrac{1}{2}\cdot \Big(-\dfrac{1}{2}\Big)\cdot \int\limits -2\cdot sin2x\cdot x\, dx +\dfrac{x^3}{6} = \\ \\ =\dfrac{1}{4}\cdot sin2x\cdot x^2+\dfrac{1}{4}\cdot \int\limits (cos2x)'\cdot x\, dx +\dfrac{x^3}{6}= \\ \\ =\dfrac{1}{4}\cdot sin2x\cdot x^2+\dfrac{1}{4}\cdot cos2x\cdot x-\dfrac{1}{4} \int\limits {cos2x} \, dx +\dfrac{x^3}{6}=$
$=\dfrac{1}{4}\cdot sin2x\cdot x^2+\dfrac{1}{4}\cdot cos2x\cdot x-\dfrac{1}{4} \cdot \dfrac{1}{2}\int\limits {(sin2x)'} \, dx +\dfrac{x^3}{6}= \\ \\= \dfrac{1}{4}\cdot sin2x\cdot x^2+\dfrac{1}{4}\cdot cos2x\cdot x-\dfrac{1}{8} \cdot sin2x +\dfrac{x^3}{6} +C\\ \\ \Rightarrow \boxed{\boxed{\int\limits {(xcosx)^2} \, dx = \dfrac{1}{4}\cdot sin2x\cdot x^2+\dfrac{1}{4}\cdot cos2x\cdot x-\dfrac{1}{8} \cdot sin2x +\dfrac{x^3}{6} +C}}$