What is the integral from 0 to pi/3 of xsin(6x) ?

The integral from 0 to pi/3 of xsin(6x) is -pi/(18)

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integral from 0 to pi/3 of xsin(6x)

\int_{0}^{\frac{π}{3}} x sin (6 x) d x

- \frac{π}{18}

Decimal Notation

- 0.17453 \dots

Solution steps

\int_{0}^{\frac{π}{3}} x sin (6 x) d x

Apply Integration By Parts

: [\frac{1}{6} (- x cos (6 x) - 6 \cdot \int - \frac{1}{6} cos (6 x) d x)]_{0}^{\frac{π}{3}}

= [\frac{1}{6} (- x cos (6 x) - 6 \cdot \int - \frac{1}{6} cos (6 x) d x)]_{0}^{\frac{π}{3}}

\int - \frac{1}{6} cos (6 x) d x = - \frac{1}{36} sin (6 x)

= [\frac{1}{6} (- x cos (6 x) - 6 (- \frac{1}{36} sin (6 x)))]_{0}^{\frac{π}{3}}

Simplify

[\frac{1}{6} (- x cos (6 x) - 6 (- \frac{1}{36} sin (6 x)))]_{0}^{\frac{π}{3}} : [\frac{1}{6} (- x cos (6 x) + \frac{1}{6} sin (6 x))]_{0}^{\frac{π}{3}}

= [\frac{1}{6} (- x cos (6 x) + \frac{1}{6} sin (6 x))]_{0}^{\frac{π}{3}}

Compute the boundaries:

- \frac{π}{18}

= - \frac{π}{18}

\int (1 + x)^{0} d x

\frac{d}{d x} (x - (\frac{1}{2})^{x})

\int cos^{3} (θ) + cos (θ) sin^{2} (θ) d θ

n = 1 \sum \infty \frac{5}{7 ^{n}}

y (x + y + 1) d x + 2 x y d y = 0

What is the integral from 0 to pi/3 of xsin(6x) ?
The integral from 0 to pi/3 of xsin(6x) is -pi/(18)