Question

алгебра 2.9(1,2)Найдите неопределенный интеграл​

Answer 1

Ответ:

$1)\ \displaystyle \int x\cdot sin^2x\, dx=\int x\cdot \dfrac{1-cos2x}{2}\, dx=\frac{1}{2}\int (x-x\cdot cos2x)\, dx=\\\\\\=\frac{1}{2}\cdot \frac{x^2}{2}-\frac{1}{2}\int x\cdot cos2x\, dx=\Big[\ u=x,\, du=dx\ ,\ dv=cos2x\, dx\ ,\ v=\dfrac{1}{2}sin2x\ \Big]=\\\\\\=\frac{x^2}{4}-\frac{1}{2}\cdot \Big(\frac{x}{2}\, sin2x-\frac{1}{2}\int sin2x\, dx\Big)=$

$\displaystyle =\frac{x^2}{4}-\frac{1}{2}\cdot \Big(\frac{x}{2}\, sin2x-\frac{1}{2}\cdot (-\frac{1}{2}\, cos2x)\Big)=\frac{x^2}{4}-\frac{x}{4}\, sin2x-\frac{1}{8}\, cos2x+C\ ;$

$2)\ \displaystyle \int x\cdot cos^2x\, dx=\int x\cdot \dfrac{1+cos2x}{2}\, dx=\frac{1}{2}\int (x+x\cdot cos2x)\, dx=\\\\\\=\frac{1}{2}\cdot \frac{x^2}{2}+\frac{1}{2}\int x\cdot cos2x\, dx=\Big[\ u=x,\, du=dx\ ,\ dv=cos2x\, dx\ ,\ v=\dfrac{1}{2}sin2x\ \Big]=\\\\\\=\frac{x^2}{4}+\frac{1}{2}\cdot \Big(\frac{x}{2}\, sin2x-\frac{1}{2}\int sin2x\, dx\Big)=$

$\displaystyle =\frac{x^2}{4}+\frac{1}{2}\cdot \Big(\frac{x}{2}\, sin2x-\frac{1}{2}\cdot (-\frac{1}{2}\, cos2x)\Big)=\frac{x^2}{4}+\frac{x}{4}\, sin2x+\frac{1}{8}\, cos2x+C\ ;$