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:
\(\begin{array}{|rcll|} \hline \int u\cdot v' &=& u\cdot \int v' - \int (u'\cdot \int v') \quad & | \quad \text{Integrate by parts}\\ \int (u'\cdot \int v') &=& u'\cdot \iint v' - \int (u''\cdot \iint v') \quad & | \quad \text{Integrate by parts} \\\\ \int u\cdot v' &=& u\cdot \int v' - \Big( u'\cdot \iint v' - \int (u''\cdot \iint v') \Big) \\ \mathbf{ \int u\cdot v' } & \mathbf{=} & \mathbf{ u\cdot \int v' - u'\cdot \iint v' + \int (u''\cdot \iint v') } \\ \hline \end{array} \)
\(\begin{array}{|llll|} \hline \int x^2 \sin^3 (x)\ dx \qquad & u = x^2 \qquad u' = 2x \qquad u'' = 2 \\ \qquad & v' = sin^3(x)\\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{ \int u\cdot v' } & \mathbf{=} & \mathbf{ u\cdot \int v' - u'\cdot \iint v' + \int (u''\cdot \iint v') } \\ \int x^2\cdot sin^3 (x) &=& x^2\cdot \int sin^3(x) - 2x\cdot \iint sin^3(x) + 2\cdot \iiint sin^3(x) \\ \hline \end{array} \)
\(\int \sin^3(x)\ dx\)
\(\begin{array}{|rcll|} \hline \int \sin^3(x) &=& \frac14 \Big( \int 3\sin(x)-\int \sin(3x) \Big) \\ \int \sin^3(x) &=& \frac14 \Big( -3\cos(x)+\frac13 \cos(3x) \Big) \\ \hline \end{array}\)
\(\iint \sin^3(x)\ dx\)
\(\begin{array}{|rcll|} \hline \iint \sin^3(x) &=& \frac14 \Big(\int -3\cos(x)+ \int \frac13 \cos(3x) \Big) \\ \iint \sin^3(x) &=& \frac14 \Big( -3\sin(x)+\frac19 \sin(3x) \Big) \\ \hline \end{array} \)
\(\iiint \sin^3(x)\ dx\)
\(\begin{array}{|rcll|} \hline \iiint \sin^3(x) &=& \frac14 \Big(\int -3\sin(x)+ \int \frac19 \sin(3x) \Big) \\ \iiint \sin^3(x) &=& \frac14 \Big( 3\cos(x)-\frac{1}{27} \cos(3x) \Big) \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \mathbf{ \int x^2\cdot \sin^3 (x) } & \mathbf{=} & \mathbf{ x^2\cdot \int sin^3(x) - 2x\cdot \iint sin^3(x) + 2\cdot \iiint sin^3(x) } \\ \int x^2\cdot \sin^3 (x) &=& x^2\cdot \frac14 \Big[ -3\cos(x)+\frac13 \cos(3x) \Big] - 2x\cdot \frac14 \Big[ -3\sin(x)+\frac19 \sin(3x) \Big] + 2\cdot \frac14 \Big[ 3\cos(x)-\frac{1}{27} \cos(3x) \Big] \\ \int x^2\cdot \sin^3 (x) &=& \frac14 x^2 \Big[ -3\cos(x)+\frac13 \cos(3x) \Big] - \frac12 x\Big[ -3\sin(x)+\frac19 \sin(3x) \Big] + \frac12 \Big[ 3\cos(x)-\frac{1}{27} \cos(3x) \Big] \\ \hline \end{array} \)
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Lol !!
