What is the integral from 0 to pi/(12) of tsin(4t) ?

The integral from 0 to pi/(12) of tsin(4t) is (-pi+3sqrt(3))/(96)

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integral from 0 to pi/(12) of tsin(4t)

\int_{0}^{\frac{π}{12}} t sin (4 t) d t

\frac{- π + 3 3}{96}

Decimal Notation

0.02140 \dots

Solution steps

\int_{0}^{\frac{π}{12}} t sin (4 t) d t

Apply Integration By Parts

: [\frac{1}{4} (- t cos (4 t) - 4 \cdot \int - \frac{1}{4} cos (4 t) d t)]_{0}^{\frac{π}{12}}

= [\frac{1}{4} (- t cos (4 t) - 4 \cdot \int - \frac{1}{4} cos (4 t) d t)]_{0}^{\frac{π}{12}}

\int - \frac{1}{4} cos (4 t) d t = - \frac{1}{16} sin (4 t)

= [\frac{1}{4} (- t cos (4 t) - 4 (- \frac{1}{16} sin (4 t)))]_{0}^{\frac{π}{12}}

Simplify

[\frac{1}{4} (- t cos (4 t) - 4 (- \frac{1}{16} sin (4 t)))]_{0}^{\frac{π}{12}} : [\frac{1}{4} (- t cos (4 t) + \frac{1}{4} sin (4 t))]_{0}^{\frac{π}{12}}

= [\frac{1}{4} (- t cos (4 t) + \frac{1}{4} sin (4 t))]_{0}^{\frac{π}{12}}

Compute the boundaries:

\frac{- π + 3 3}{96}

= \frac{- π + 3 3}{96}

n = 0 \sum \infty \frac{1}{n + 12}

\frac{\partial}{\partial x} (4 x^{2} y - 2 x + 2 y^{3})

\int_{0}^{\frac{π}{4}} 4 tan^{4} (x) d x

y = sin (sin (x)), (2 π, 0)

\int 7 x cos (2 x) d x

What is the integral from 0 to pi/(12) of tsin(4t) ?
The integral from 0 to pi/(12) of tsin(4t) is (-pi+3sqrt(3))/(96)