What is the integral from 0 to pi/4 of xsin(12x) ?

The integral from 0 to pi/4 of xsin(12x) is pi/(48)

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integral from 0 to pi/4 of xsin(12x)

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

\frac{π}{48}

Decimal

0.06544 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{1}{12} cos (12 x) d x = - \frac{1}{144} sin (12 x)

= [\frac{1}{12} (- x cos (12 x) - 12 (- \frac{1}{144} sin (12 x)))]_{0}^{\frac{π}{4}}

Simplify

[\frac{1}{12} (- x cos (12 x) - 12 (- \frac{1}{144} sin (12 x)))]_{0}^{\frac{π}{4}} : [\frac{1}{12} (- x cos (12 x) + \frac{1}{12} sin (12 x))]_{0}^{\frac{π}{4}}

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

Compute the boundaries:

\frac{π}{48}

= \frac{π}{48}

\int \frac{4}{3 x ^{2}} d x

y^{^{''}} - 4 y^{^{'}} - 5 y = - 2, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (e^{3 x} cos (x))

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

\int (π - x) cos (π x) d x

What is the integral from 0 to pi/4 of xsin(12x) ?
The integral from 0 to pi/4 of xsin(12x) is pi/(48)