Notes: How do you find the integral of e^(-x)sin(2x)dx using integration by parts? Here's my work: u = sin(2x) dv to be e^(-x)dx du = 2cos(2x)dx v =...

Saturday, 28 September 2013

How do you find the integral of e^(-x)sin(2x)dx using integration by parts? Here's my work: u = sin(2x) dv to be e^(-x)dx du = 2cos(2x)dx v =...

You need
to avoid the perpetual cycle, hence, you should use the following notation $\displaystyle\int$ $\displaystyle{{e}}^{{-{x}}}{\sin{{2}}}{x}$
dx $\displaystyle={I}{\quad\text{and}\quad}{y}{o}{u}\ne{e}{d}\to{u}{s}{e}{p}{a}{r}{t}{s}\to{s}{o}{l}{v}{e}{t}{h}{e}\int{e}{g{{r}}}{a}{l}{s}{u}{c}{h}{t}{\hat{:}}$

$\displaystyle{u}=$
e^(-x) => du = -e^(-x)dx

$\displaystyle{d}{v}={\sin{{2}}}{x}{\left.{d}{x}\right.}\Rightarrow{v}=-\frac{{{\cos{{2}}}{x}}}{{2}}$

Using the following formula yields:

$\displaystyle\int{u}{d}{v}={u}{v}-\int$
vdu

$\displaystyle\int{{e}}^{{-{x}}}{\sin{{2}}}{x}{\left.{d}{x}\right.}=-\frac{{{{e}}^{{-{x}}}\cdot{\cos{{2}}}{x}}}{{2}}-{\left(\frac{{1}}{{2}}\right)}\int{{e}}^{{-{x}}}{\cos{{2}}}{x}{\left.{d}{x}\right.}$

You should solve the integral $\displaystyle\int{{e}}^{{-{x}}}{\cos{{2}}}{x}{\left.{d}{x}\right.}$ using parts such
that:

$\displaystyle{u}=\ldots$

Notes

Saturday, 28 September 2013

How do you find the integral of e^(-x)sin(2x)dx using integration by parts? Here's my work: u = sin(2x) dv to be e^(-x)dx du = 2cos(2x)dx v =...

No comments:

Post a Comment

In 1984, is Julia a spy? Please provide specific examples from the book. My teacher says that he knows of 17 pieces of evidence which proves that Julia...