What is the integral from 0 to pi of xsin(ax) ?

The integral from 0 to pi of xsin(ax) is (sin(pia)-piacos(pia))/(a^2)

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integral from 0 to pi of xsin(ax)

\int_{0}^{π} x sin (a x) d x

\frac{sin ( πa ) - πa cos ( πa )}{a ^{2}}

Solution steps

\int_{0}^{π} x sin (a x) d x

Apply Integration By Parts

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

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

\int - \frac{1}{a} cos (a x) d x = - \frac{1}{a ^{2}} sin (a x)

= [\frac{1}{a} (- x cos (a x) - a (- \frac{1}{a ^{2}} sin (a x)))]_{0}^{π}

Simplify

[\frac{1}{a} (- x cos (a x) - a (- \frac{1}{a ^{2}} sin (a x)))]_{0}^{π} : [\frac{1}{a} (- x cos (a x) + \frac{1}{a} sin (a x))]_{0}^{π}

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

Compute the boundaries:

\frac{1}{a} (\frac{1}{a} sin (πa) - π cos (πa))

= \frac{1}{a} (\frac{1}{a} sin (πa) - π cos (πa))

Simplify

= \frac{sin ( πa ) - πa cos ( πa )}{a ^{2}}

\int_{0}^{\frac{5 π}{3}} 4 tan (\frac{x}{5}) d x

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\int_{1}^{4} \frac{5}{x} - \frac{5}{x ^{2}} d x

\int_{0}^{40} 0.7 x d x

What is the integral from 0 to pi of xsin(ax) ?
The integral from 0 to pi of xsin(ax) is (sin(pia)-piacos(pia))/(a^2)