Problems on Techniques of Integration

Use the Integration by Parts. Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=& x^2\\
dv &=& \sin(x) dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&2 x dx\\
v &=& -\cos(x)\;.
\end{array}\right.\end{displaymath}

So

\begin{displaymath}\int x^2 \sin(x) dx = - x^2 \cos(x) + \int 2 x \cos(x) dx = - x^2 \cos(x) + 2 \int x \cos(x) dx\;.\end{displaymath}

In order to integrate the function $x \cos(x)$, we will need to do another integration by parts. Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=&x\\
dv &=& \cos(x)dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&dx\\
v &=& \sin(x)\;.
\end{array}\right.\end{displaymath}

So

\begin{displaymath}\int x \cos(x) dx = x \sin(x) - \int \sin(x) dx = x \sin(x) + \cos(x)\;.\end{displaymath}

Hence

\begin{displaymath}\int x^2 \sin(x) dx = - x^2 \cos(x) + x \sin(x) + \cos(x) + C\;.\end{displaymath}

Detailed Answer.


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