Question

Calculați integralele folosind metoda integrării prin părți :

$1) \int {x}^{2} \: arctgx \: dx \: ,x \: \in \: \mathbb{R}$

$2) \int \: x \: arcsinx \: dx \: , \: x \: \in \: (-1,1)$

$3) \int ln(x + \sqrt{1 + {x}^{2} } ) \: dx \: , \: x \: \in \: (0, + \infty )$

$4) \int x \: {sin}^{2} x \: dx \: ,x \: \in \: \mathbb{R}$

$5) \int x \: {cos}^{2} x \: dx \: ,x \: \in \: \mathbb{R}$

Answer 1

$2J=x\cdot \sqrt{1-x^2}+\arcsin x\Rightarrow J=\boxed{\dfrac{x\cdot \sqrt{1-x^2}}{2}+\dfrac{\arcsin x}{2}}\\\\\text{Revenind,obtinem:} \dfrac{x^2\cdot \arcsin x}{2}+\dfrac{x\sqrt{1-x^2}}{4}+\dfrac{\arcsin x}{4}-\dfrac{\arcsin x}{2}+C$$a) \displaystyle \int x^2 \cdot arctg x dx\\f '=x^2,g=arctg x \left(\rightarrow f=\dfrac{x^3}{3},g'=\dfrac{1}{x^2+1} \right)\\\int x^2 \cdot arctg x=\dfrac{x^3}{3}\cdot arctg x-\int \dfrac{x^3}{3(x^2+1)}dx =\dfrac{x^3}{3}\cdot arctg x-\\\dfrac{1}{3}\cdot \int \dfrac{x^3+x-x}{x^2+1} dx=\dfrac{x^3}{3}\cdot arctg x-\dfrac{1}{3} \int \dfrac{x(x^2+1)}{x^2+1} dx +\\\dfrac{1}{3} \int \dfrac{x}{x^2+1} dx=\dfrac{x^3}{3}\cdot arctg x-\dfrac{1}{3}\int x dx+\dfrac{1}{3}\cdot \dfrac{1}{2}\cdot \ln(x^2+1)=$

$=\dfrac{x^3}{3}-\dfrac{x^2}{6}+\dfrac{\ln(x^2+1)}{6} +C$

$2)\displaystyle \int x\cdot \arcsin x dx\\\\f' =x ,g =arcsin x \left(\rightarrow f=\dfrac{x^2}{2} , g'=\dfrac{1}{\sqrt{1-x^2}}\right)\\\int x\cdot arcsin x=\int \left(\dfrac{x^2}{2}\right)'\cdot arcsin x dx=\dfrac{x^2}{2}\cdot \arcsin x-\\\dfrac{1}{2}\cdot \int \dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^2}{2}\cdot \arcsin x-\dfrac{1}{2}\cdot \int \dfrac{x^2-1+1}{\sqrt{1-x^2}} dx=\\\dfrac{x^2}{2}\cdot arcsin x+\dfrac{1}{2}\int \dfrac{1-x^2}{\sqrt{1-x^2}}dx -\dfrac{1}{2}\cdot \int \dfrac{1}{\sqrt{1-x^2}} dx=$

$\displaystyle \\\dfrac{x^2}{2}\cdot \arcsin x +\dfrac{1}{2} \int \sqrt{1-x^2} dx -\dfrac{\arcsin x}{2}.\\\text{Rezolvam separat:}\\Fie J=\int \sqrt{1-x^2}dx=\int (x')\cdot \sqrt{1-x^2} dx=x\cdot \sqrt{1-x^2}-\int \dfrac{x^2}{\sqrt{1-x^2}} dx\\=x\cdot \sqrt{1-x^2}+\int \dfrac{x^2-1+1}{\sqrt{1-x^2}} dx=x\cdot \sqrt{1-x^2}-\int \sqrt{1-x^2} dx +\\\int \dfrac{1}{\sqrt{1-x^2}} dx=x\cdot \sqrt{1-x^2}-J+\arcsin x.$

$4)+5) \displaystyle \text{Fie }I=\int x\cdot \sin^2 x dx\text{ si }J=\int x\cdot \cos^2 xdx\\I+J=\int x(\sin^2x+\cos^2 x) dx=\int x dx =\dfrac{x^2}{2}+C\\J-I=\int x(\cos^2 x-\sin^2 x)dx =\int x\cdot \cos 2x dx =\dfrac{1}{2} \int x \cdot (\sin 2x)' dx=\\=\dfrac{x\cdot \sin 2x}{2}-\dfrac{1}{2}\int \sin 2x dx=\dfrac{x\cdot \sin 2x}{2}+\dfrac{\cos 2x}{4}+C\\\text{Rezolvand sistemul obtinem:}\boxed{J=\dfrac{x^2}{4}+\dfrac{x\cdot \sin 2x}{4}+\dfrac{\cos 2x}{8}+C}$

$\boxed{I=\dfrac{x^2}{2}-\dfrac{x\cdot \sin2 x}{4} -\dfrac{\cos 2x}{8}+C}$