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 January 24th, 2013, 12:58 AM #1 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 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!
 January 24th, 2013, 01:06 AM #2 Math Team   Joined: Apr 2010 Posts: 2,702 Thanks: 331 Re: int x^2sin2x dx Can you do it using integration by parts? It was shown to you here.
 January 24th, 2013, 02:19 AM #3 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 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!
 January 24th, 2013, 02:25 AM #4 Math Team   Joined: Apr 2010 Posts: 2,702 Thanks: 331 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 $2\int \sin(x)\cos(x) dx= 2\int \sin(x) d\sin(x)$
 January 24th, 2013, 06:46 AM #5 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: int x^2sin2x dx Hello! so:$\int x^2sin2x=?$ $u=x^2$ $du=2$ $dv=sin2x$ $v=(-\frac{1}{2}cosx$ so: $=uv-\int vdu$ $=x^2(-\frac{1}{2}cosx)-\int(-\frac{1}{2}cosx)( 2)$ [latex]=(-\frac{x^2}{2}cosx-\int((-cosx)[latex] we should get: $\frac{1}{4}(1-2x)xos2x+\frac{1}{2}xsin2x+C$ Can someone help? Many thanks!!!
 January 24th, 2013, 07:57 AM #6 Math Team   Joined: Apr 2010 Posts: 2,702 Thanks: 331 Re: int x^2sin2x dx u = x^2 yields du = 2x dx instead of 2. Can you get it from here?
 January 24th, 2013, 08:08 AM #7 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: int x^2sin2x dx [color=#0000FF]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? [/color]
January 24th, 2013, 08:28 AM   #8
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,266
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Re: int x^2sin2x dx

Hello, ungeheuer!

Quote:
 $I \;=\;\int x^2\,\!\sin2x\,dx$ Please, the result is 1/4[color=beige] . [/color][color=blue]How?[/color]

Integrate by parts:

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

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

By parts again:

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

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

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

 January 24th, 2013, 09:15 AM #9 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: int x^2sin2x dx [b]Many thanks, soroban! All clear now!!!

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