What is the integral from 0 to pi/6 of tsin(2t) ?

The integral from 0 to pi/6 of tsin(2t) is (-pi+3sqrt(3))/(24)

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integral from 0 to pi/6 of tsin(2t)

\int_{0}^{\frac{π}{6}} t sin (2 t) d t

\frac{- π + 3 3}{24}

Decimal Notation

0.08560 \dots

Solution steps

\int_{0}^{\frac{π}{6}} t sin (2 t) d t

Apply Integration By Parts

: [\frac{1}{2} (- t cos (2 t) - 2 \cdot \int - \frac{1}{2} cos (2 t) d t)]_{0}^{\frac{π}{6}}

= [\frac{1}{2} (- t cos (2 t) - 2 \cdot \int - \frac{1}{2} cos (2 t) d t)]_{0}^{\frac{π}{6}}

\int - \frac{1}{2} cos (2 t) d t = - \frac{1}{4} sin (2 t)

= [\frac{1}{2} (- t cos (2 t) - 2 (- \frac{1}{4} sin (2 t)))]_{0}^{\frac{π}{6}}

Simplify

[\frac{1}{2} (- t cos (2 t) - 2 (- \frac{1}{4} sin (2 t)))]_{0}^{\frac{π}{6}} : [\frac{1}{2} (- t cos (2 t) + \frac{1}{2} sin (2 t))]_{0}^{\frac{π}{6}}

= [\frac{1}{2} (- t cos (2 t) + \frac{1}{2} sin (2 t))]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{- π + 3 3}{24}

= \frac{- π + 3 3}{24}

\frac{4}{s ^{2} ( s + 2 )} + \frac{3}{s + 2}

\frac{d}{d x} (arctan (\frac{x}{4 - 16 - x ^{2}}))

x \to 1 lim (\frac{2 x}{6 x - 10})

\frac{d}{d x} (60)

x x (9.27)

What is the integral from 0 to pi/6 of tsin(2t) ?
The integral from 0 to pi/6 of tsin(2t) is (-pi+3sqrt(3))/(24)