Problems on Techniques of Integration

The trigonometric functions $\cos$ and $\sin$ are nice functions since their derivatives and antiderivatives are easy to obtain. So for the integration by parts, these functions have the same behavior whether we differentiate them or take their antiderivatives. Therefore the focus should be on the other function involved in the integration. In this case, we must differentiate $f(x) = x$ because its derivative gives the constant 1.

Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=&x\\
dv &=& \sin(3x)dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&dx\\
v &=&\displaystyle - \frac{1}{3}\cos(x)\;.
\end{array}\right.\end{displaymath}

Since

\begin{displaymath}\int u dv = u v - \int v du\;,\end{displaymath}

we get which implies

\begin{displaymath}\int x \sin(3 x) dx = - x \frac{1}{3}\cos(x) - \int - \frac{1}{3}\cos(x) dx \end{displaymath}

or

\begin{displaymath}\int x \sin(3 x) dx =- \frac{x}{3}\cos(x) + \frac{1}{3}\sin(x) + C \;.\end{displaymath}

It is a common mistake to forget the constant $C$.


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