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

The integral from 0 to pi/3 of xsin(9x) is pi/(27)

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

\int_{0}^{\frac{π}{3}} x sin (9 x) d x

\frac{π}{27}

Decimal

0.11635 \dots

Solution steps

\int_{0}^{\frac{π}{3}} x sin (9 x) d x

Apply Integration By Parts

: [\frac{1}{9} (- x cos (9 x) - 9 \cdot \int - \frac{1}{9} cos (9 x) d x)]_{0}^{\frac{π}{3}}

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

\int - \frac{1}{9} cos (9 x) d x = - \frac{1}{81} sin (9 x)

= [\frac{1}{9} (- x cos (9 x) - 9 (- \frac{1}{81} sin (9 x)))]_{0}^{\frac{π}{3}}

Simplify

[\frac{1}{9} (- x cos (9 x) - 9 (- \frac{1}{81} sin (9 x)))]_{0}^{\frac{π}{3}} : [\frac{1}{9} (- x cos (9 x) + \frac{1}{9} sin (9 x))]_{0}^{\frac{π}{3}}

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

Compute the boundaries:

\frac{π}{27}

= \frac{π}{27}

\frac{1}{( s ^{2} + 1 ) ^{2}}

\frac{x ^{2}}{7 + 8 x}

\frac{d}{d x} (3 x^{4} (4 - x)^{3})

\int \frac{8}{( 1 + 4 t ) ^{2}} d t

\frac{d}{d x} (8 + x)

What is the integral from 0 to pi/3 of xsin(9x) ?
The integral from 0 to pi/3 of xsin(9x) is pi/(27)