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Integrate the following:  ∫x^2 sin^3 (x) dx. Also, please show steps of solution.

I thank you for any help.

Guest Apr 2, 2017
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9+0 Answers

 #3
avatar+26365 
+2

First part:

Alan  Apr 2, 2017
 #4
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+2

Second part:

Alan  Apr 2, 2017
 #5
avatar+26365 
+3

Third part:

Alan  Apr 2, 2017
 #6
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+2

Thank you Alan.

Guest Apr 2, 2017
 #7
avatar+91226 
+3

Integrate the following:  ∫x^2 sin^3 (x) dx. Also, please show steps of solution.

I thank you for any help.

 

\(\int\;x^2sin^3x\;dx\\ =\int\;x^2sinx\;sin^2x\;dx\\ =\int\;x^2sinx\;sin^2x\;dx\\ \qquad \qquad cos2x=cos^2x-sin^2x\\ \qquad \qquad cos2x=1-2sin^2x\\ \qquad \qquad 2sin^2x=1-cos2x\\ \qquad \qquad sin^2x=\frac{1-cos2x}{2}\\ =\int\;x^2sinx\;\frac{1-cos2x}{2}\;dx\\ =\int\;\;\frac{x^2sin(x)-x^2sinxcos(2x)}{2}\;dx\\ =\frac{-1}{2}\int\;[x^2sin(x)cos(2x)-x^2sin(x)]\;dx\\ \)

 

\(\qquad \qquad sin(\alpha-\beta)+sin(\alpha+\beta)\\ \qquad \qquad =sin(\alpha)cos(\beta)-sin(\beta)cos(\alpha)+sin(\alpha)cos(\beta)+sin(\beta)cos(\alpha)\\ \qquad \qquad =2sin(\alpha)cos(\beta)\\ \qquad \qquad so\\ \qquad \qquad 2sin(x)cos(2x)=sin(x-2x)+sin(x+2x)\\ \qquad \qquad 2sin(x)cos(2x)=sin(-x)+sin(3x)\\ \qquad \qquad 2sin(x)cos(2x)=sin(3x)-sin(x)\\ \qquad \qquad sin(x)cos(2x)=\frac{1}{2}[sin(3x)-sin(x)]\\ =\frac{-1}{2}\int\;[x^2sin(x)cos(2x)-x^2sin(x)]\;dx\\ =\frac{-1}{2}\int\;\frac{x^2}{2}[sin(3x)-sin(x)]-x^2sin(x)]\;dx\\ =\frac{-1}{2}\int\;[\frac{x^2}{2}sin(3x)-\frac{x^2}{2}sin(x)-x^2sin(x)]\;dx\\ =\frac{-1}{2}\int\;[\frac{x^2}{2}sin(3x)-\frac{3x^2}{2}sin(x)]\;dx\\ =\frac{-1}{4}\left(\int\;[x^2sin(3x)]dx\;-\;\int[3x^2sin(x)]\;dx\right)\\ \text{Now work out each of these two integral seperately using integration by parts.}\)

 

Maybe I will finish it later  (it is 1:36am)  anyway, that is a good start for you :)

Melody  Apr 2, 2017
 #8
avatar+91226 
+1

I see Alan beat me to it :))

Melody  Apr 2, 2017
 #9
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+1

Thank you Melody. I shall take it from here.

Guest Apr 2, 2017
 #10
avatar+26365 
+2

Here's an alternative approach:

 

Alan  Apr 3, 2017
 #11
avatar+18777 
+2

Integrate the following:  ∫x^2 sin^3 (x) dx. Also, please show steps of solution.

I thank you for any help.

 

Formula \(\sin^3(x)\):

\(\begin{array}{|rcll|} \hline \sin(3x) &=& \sin(2x+x) \\ &=& \underbrace{ \sin(2x) }_{=2\sin(x)\cos(x)}\cdot \cos(x)+\underbrace{ \cos(2x) }_{=\cos^2(x)-\sin^2(x)}\cdot \sin(x) \\ &=& 2\sin(x)\cos^2(x) + \cos^2(x)\sin(x)-\sin^3(x) \\ &=& 3\sin(x)\cos^2(x) -\sin^3(x) \\ &=& 3\sin(x) \Big( 1-\sin^2(x) \Big) -\sin^3(x) \\ &=& 3\sin(x) -3\sin^3(x) -\sin^3(x) \\ \sin(3x)&=& 3\sin(x) -4\sin^3(x) \\\\ 4\sin^3(x) &=& 3\sin(x) - \sin(3x) \\ \mathbf{ \sin^3(x) } & \mathbf{=} & \mathbf{ \frac14 \Big( 3\sin(x) - \sin(3x) \Big) } \\ \hline \end{array}\)

 

Double Integration by parts:

\(\begin{array}{|rcll|} \hline \int u\cdot v' &=& u\cdot \int v' - \int (u'\cdot \int v') \quad & | \quad \text{Integrate by parts}\\ \int (u'\cdot \int v') &=& u'\cdot \iint v' - \int (u''\cdot \iint v') \quad & | \quad \text{Integrate by parts} \\\\ \int u\cdot v' &=& u\cdot \int v' - \Big( u'\cdot \iint v' - \int (u''\cdot \iint v') \Big) \\ \mathbf{ \int u\cdot v' } & \mathbf{=} & \mathbf{ u\cdot \int v' - u'\cdot \iint v' + \int (u''\cdot \iint v') } \\ \hline \end{array} \)

 

\(\begin{array}{|llll|} \hline \int x^2 \sin^3 (x)\ dx \qquad & u = x^2 \qquad u' = 2x \qquad u'' = 2 \\ \qquad & v' = sin^3(x)\\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{ \int u\cdot v' } & \mathbf{=} & \mathbf{ u\cdot \int v' - u'\cdot \iint v' + \int (u''\cdot \iint v') } \\ \int x^2\cdot sin^3 (x) &=& x^2\cdot \int sin^3(x) - 2x\cdot \iint sin^3(x) + 2\cdot \iiint sin^3(x) \\ \hline \end{array} \)

 

 

\(\int \sin^3(x)\ dx\)

\(\begin{array}{|rcll|} \hline \int \sin^3(x) &=& \frac14 \Big( \int 3\sin(x)-\int \sin(3x) \Big) \\ \int \sin^3(x) &=& \frac14 \Big( -3\cos(x)+\frac13 \cos(3x) \Big) \\ \hline \end{array}\)

 

\(\iint \sin^3(x)\ dx\)

\(\begin{array}{|rcll|} \hline \iint \sin^3(x) &=& \frac14 \Big(\int -3\cos(x)+ \int \frac13 \cos(3x) \Big) \\ \iint \sin^3(x) &=& \frac14 \Big( -3\sin(x)+\frac19 \sin(3x) \Big) \\ \hline \end{array} \)

 

\(\iiint \sin^3(x)\ dx\)

\(\begin{array}{|rcll|} \hline \iiint \sin^3(x) &=& \frac14 \Big(\int -3\sin(x)+ \int \frac19 \sin(3x) \Big) \\ \iiint \sin^3(x) &=& \frac14 \Big( 3\cos(x)-\frac{1}{27} \cos(3x) \Big) \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ \int x^2\cdot \sin^3 (x) } & \mathbf{=} & \mathbf{ x^2\cdot \int sin^3(x) - 2x\cdot \iint sin^3(x) + 2\cdot \iiint sin^3(x) } \\ \int x^2\cdot \sin^3 (x) &=& x^2\cdot \frac14 \Big[ -3\cos(x)+\frac13 \cos(3x) \Big] - 2x\cdot \frac14 \Big[ -3\sin(x)+\frac19 \sin(3x) \Big] + 2\cdot \frac14 \Big[ 3\cos(x)-\frac{1}{27} \cos(3x) \Big] \\ \int x^2\cdot \sin^3 (x) &=& \frac14 x^2 \Big[ -3\cos(x)+\frac13 \cos(3x) \Big] - \frac12 x\Big[ -3\sin(x)+\frac19 \sin(3x) \Big] + \frac12 \Big[ 3\cos(x)-\frac{1}{27} \cos(3x) \Big] \\ \hline \end{array} \)

 

laugh

heureka  Apr 4, 2017

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