## int x^2sin2x dx

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### int x^2sin2x dx

I am given to evaluate:

$\int x^2sin(2x)\,dx$

Please, the result is 1/4...I can't get this...should it be just normal multiplying rule for integrals or something else?

Many thanks!
Last edited by MarkFL on Thu Jan 24, 2013 1:14 pm, edited 1 time in total.
Reason: Fix LaTeX and typos
"Blessed are the poor in spirit, for theirs is the kingdom of heaven." Matthew 5
Intelligence is the ability of a man to be happy. (Seneca)
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ungeheuer
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### Re: int x^2sin2x dx

Can you do it using integration by parts? It was shown to you here.
Hoempa
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### Re: int x^2sin2x dx

Hello!
The problem is:

dv= sin2x

how do we get v then?
$\frac{d}{dx}=sin2x$
so we search what would give us derivative =sin2x, am I right?
so sin2x is derivative of v
And that is problem!
Many thanks!
"Blessed are the poor in spirit, for theirs is the kingdom of heaven." Matthew 5
Intelligence is the ability of a man to be happy. (Seneca)
Cute people learn form stupid ones (A. Schopenhauer)

ungeheuer
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### Re: int x^2sin2x dx

One way is to use [-cos(a)]' = sin(a) * [a]'.
Sub 2x = a in your equation.

Another is use $\sin(2x) = 2\sin(x)\cos(x)$ so $2\int \sin(x)\cos(x) dx = 2\int \sin(x) d\sin(x)$
Hoempa
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### Re: int x^2sin2x dx

Hello!
so:$\int x^2sin2x=?$
$u=x^2$
$du=2$
$dv=sin2x$
$v=(-\frac{1}{2}cosx$

so:
$=uv-\int vdu$
$=x^2(-\frac{1}{2}cosx)-\int(-\frac{1}{2}cosx)( 2)$
$=(-\frac{x^2}{2}cosx-\int((-cosx)[latex]
we should get:

[latex]\frac{1}{4}(1-2x)xos2x+\frac{1}{2}xsin2x+C$

Can someone help?
Many thanks!!!
"Blessed are the poor in spirit, for theirs is the kingdom of heaven." Matthew 5
Intelligence is the ability of a man to be happy. (Seneca)
Cute people learn form stupid ones (A. Schopenhauer)

ungeheuer
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### Re: int x^2sin2x dx

u = x^2 yields du = 2x dx instead of 2. Can you get it from here?
Hoempa
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### Re: int x^2sin2x dx

Huh!
$u=x^2$
$du=2dx$
$dv=sin2x$
$v=(-\frac{1}{2}cos2x)$

$result=x^2(-\frac{1}{2}cos2x)-\int (-\frac{1}{2}cos2x)(2dx)=(-\frac{x^2}{2}cos2x)+\int (cos2x)=(-\frac{x^2}{2}cos2x)-\frac{1}{2}sin2x+C)=(-\frac{1}{2}(x^2cos2x+sin2x)+C$

but the result is up there... not that...please, where is the mistake?

"Blessed are the poor in spirit, for theirs is the kingdom of heaven." Matthew 5
Intelligence is the ability of a man to be happy. (Seneca)
Cute people learn form stupid ones (A. Schopenhauer)

ungeheuer
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### Re: int x^2sin2x dx

Hello, ungeheuer!

$I \;=\;\int x^2\,\!\sin2x\,dx$

Please, the result is 1/4 . How?

Integrate by parts:

. . $\begin{Bmatrix} u &=& x^2 &\;\;& dv &=& \sin2x\,dx \\ \\ \\ du &=& 2x\,dx && v &=& -\frac{1}{2}\cos2x \end{Bmatrix}$

$I \;=\;-\frac{1}{2}x^2\cos2x\,+\,\int x\,\!\cos2x\,dx$

By parts again:

. . $\begin{Bmatrix}u &=& x &\;\;& dv &=& \cos2x\,dx \\ \\ \\ du &=& dx && v &=& \frac{1}{2}\sin2x \end{Bmatrix}$

$I \;=\;-\frac{1}{2}x^2\cos2x\,+\,\left[\frac{1}{2}x\sin2x \,-\,\frac{1}{2}\int\sin2x\,dx\right]$

$I \;=\;-\frac{1}{2}x^2\cos2x \.+\,\frac{1}{2}x\sin2x \,+\,\frac{1}{4}\cos2x \,+\,C$

soroban
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### Re: int x^2sin2x dx

[b]Many thanks, soroban!
All clear now!!!
"Blessed are the poor in spirit, for theirs is the kingdom of heaven." Matthew 5
Intelligence is the ability of a man to be happy. (Seneca)
Cute people learn form stupid ones (A. Schopenhauer)

ungeheuer
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