What is the integral from 0 to pi/3 of xsin(3x) ?

The integral from 0 to pi/3 of xsin(3x) is pi/9

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integral from 0 to pi/3 of xsin(3x)

\int_{0}^{\frac{π}{3}} x sin (3 x) d x

\frac{π}{9}

Decimal

0.34906 \dots

Solution steps

\int_{0}^{\frac{π}{3}} x sin (3 x) d x

Apply Integration By Parts

: [\frac{1}{3} (- x cos (3 x) - 3 \cdot \int - \frac{1}{3} cos (3 x) d x)]_{0}^{\frac{π}{3}}

= [\frac{1}{3} (- x cos (3 x) - 3 \cdot \int - \frac{1}{3} cos (3 x) d x)]_{0}^{\frac{π}{3}}

\int - \frac{1}{3} cos (3 x) d x = - \frac{1}{9} sin (3 x)

= [\frac{1}{3} (- x cos (3 x) - 3 (- \frac{1}{9} sin (3 x)))]_{0}^{\frac{π}{3}}

Simplify

[\frac{1}{3} (- x cos (3 x) - 3 (- \frac{1}{9} sin (3 x)))]_{0}^{\frac{π}{3}} : [\frac{1}{3} (- x cos (3 x) + \frac{1}{3} sin (3 x))]_{0}^{\frac{π}{3}}

= [\frac{1}{3} (- x cos (3 x) + \frac{1}{3} sin (3 x))]_{0}^{\frac{π}{3}}

Compute the boundaries:

\frac{π}{9}

= \frac{π}{9}

\int_{- 1}^{2} (3 x + 6) d x

\int_{- 1}^{1} \frac{x ^{4} + 2}{x ^{2} + 1} d x

\int_{1}^{5} (\frac{3 x ^{2} + 6 x + 1}{x}) d x

\int_{0}^{\frac{π}{2}} x d x

\int_{0}^{1} (x) - (x^{2}) d x

What is the integral from 0 to pi/3 of xsin(3x) ?
The integral from 0 to pi/3 of xsin(3x) is pi/9