tex]...

April 2021 1 24 Report

[tex] \int\limits^1_0 { x^{2} cos(nx)} \, dx \\ \int\limits^1_0 { x^{2} sin(nx)} \, dx \\ \int\limits^ \pi _1 { \frac{2 \pi -2x}{\pi -1} cos(nx) } \, dx \\ \int\limits^ \pi _1 { \frac{2 \pi -2x}{\pi -1} sin(nx) } \, dx [/tex]

Answers & Comments

NNNLLL54

Verified answer

$\int _0^1x^2cos(nx)dx=\\\\=[u=x^2,\; du=2x\, dx,\; dv=cos(nx)dx,\; v=\frac{1}{n}sin(nx)]=\\\\=\frac{x^2}{n}sin(nx)|_0^1-\frac{2}{n}\int _0^1x\cdot sin(nx)dx=\\\\=[u=x,\; du=dx,\; dv=sin(nx)dx,\; v=-\frac{1}{n}cos(nx)]=\\\\=\frac{1}{n}sinn-\frac{2}{n}\cdot (-\frac{x}{n}cos(nx)|_0^1+\frac{1}{n}\int _0^1cos(nx)dx)=\\\\=\frac{1}{n}sinn-\frac{2}{n}(-\frac{1}{n}cosn+\frac{1}{n^2}sin(nx)|_0^1)=\\\\=\frac{1}{n}sinn+\frac{2}{n^2}cosn-\frac{2}{n^3}sinn$

$2)\; \; \int _1^{\pi } \frac{2\pi -2x}{\pi -1} cos(nx)dx=\\\\=[u=\frac{2\pi -2x}{\pi -1},\; du=-\frac{2}{\pi -1}dx,\; v=\frac{1}{n}sin(nx)]=\\\\=\frac{2(\pi -x)}{\pi -1}\cdot \frac{1}{n}sin(nx)|_1^{\pi }+\frac{2}{n(\pi -1)}\int _1^{\pi }sin(nx)dx=\\\\=-\frac{2}{n}sin\, n-\frac{2}{n(\pi -1)}\cdot \frac{1}{n}cos(nx)|_1^{\pi }=\\\\=-\frac{2}{n}sin\, n-\frac{2}{n^2(\pi -1)}\cdot (cos(\pi n)-cos\, n)=\\\\=-\frac{2}{n}sin\, n-\frac{2}{n^2(\pi -1)}\cdot ((-1)^{n}-cos\, n)$

2 votes Thanks 1

pomogite pozhalujsta reshit hot chto to9244456225f04bcb35f87a6271f639b6 35274

Answer

Answers & Comments

Verified answer

pomogite pozhalujsta reshit hot chto to9244456225f04bcb35f87a6271f639b6 35274

рекомендуемые вопросы

Helpful Links

Helpful Social