Problems on Techniques of Integration

Use the Integration by Parts. Set

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

Then

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

So

$$\int x \sin(3 x) dx = - x \frac{1}{3}\cos(x) - \int - \frac{1}{3}\cos(x) dx$$

or

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

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