How do you integrate $\int t sin (m t) d t$ $int t sin(mt)dt$ where m does not equal 0?

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Answer 1 · 2018-03-19T14:37:50

$\frac{sin (m t) - m t cos (m t)}{m^{2}} + C$ $(sin(mt)-mtcos(mt))/m^2+C$

We have:

$\int t sin (m t) d t$ $inttsin(mt)dt$ , where $m$ $m$ is presumably real, and does not equal zero. We say that $m \neq 0$ $m!=0$ and $m inRR$ .

According to integration by parts, $intf(x)g(x)dx=f(x)intg(x)dx-intf'(x)(intg(x)dx)dx$

Here, $f(t)=t$ and $g(t)=sin(mt)$ . We have:

$tintsin(mt)dt-int(intsin(mt)dt)dt$

We need to compute $intsin(mt)dt$ . Using integration by substitution,

$intf(g(x))g'(x)dx=intf(u)du$ , where $u=g(x)$ . Here, $u=mt$ , and $u'=m$ , so we can say:

$1/m intsin(mt)mdt$

$1/m intsin(u)du$

$1/m*-cos(u)$

$-cos(mt)/m$ . Inputting:

$-(tcos(mt))/m-int-cos(mt)/mdt$

$-(tcos(mt))/m+1/m intcos(mt)dt$

Using integration by substitution, where $u=mt$ again, we have:

$-(tcos(mt))/m+1/m(1/m intcos(mt)mdt)$

$-(tcos(mt))/m+1/m(1/m intcos(u)du)$

$-(tcos(mt))/m+1/m(1/m sin(mt))$

$sin(mt)/m^2-(tcos(mt))/m$

$(sin(mt)-mtcos(mt))/m^2+C$

How do you integrate int t sin(mt)dt∫tsin(mt)dtint t sin(mt)dt where m does not equal 0?