What is the integral from 0 to pi/2 of xsin^2(x) ?

The integral from 0 to pi/2 of xsin^2(x) is (pi^2+2)/(16)+1/8

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integral from 0 to pi/2 of xsin^2(x)

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

\frac{π ^{2} + 2}{16} + \frac{1}{8}

Decimal

0.86685 \dots

Solution steps

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

Apply Integration By Parts

= [\frac{1}{4} (x (2 x - sin (2 x)) - 4 \cdot \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x)]_{0}^{\frac{π}{2}}

\int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x = \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))

= [\frac{1}{4} (x (2 x - sin (2 x)) - 4 \cdot \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x)))]_{0}^{\frac{π}{2}}

Simplify

[\frac{1}{4} (x (2 x - sin (2 x)) - 4 \cdot \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x)))]_{0}^{\frac{π}{2}} : [\frac{1}{8} (2 x^{2} - 2 x sin (2 x) - cos (2 x))]_{0}^{\frac{π}{2}}

= [\frac{1}{8} (2 x^{2} - 2 x sin (2 x) - cos (2 x))]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π ^{2} + 2}{16} + \frac{1}{8}

= \frac{π ^{2} + 2}{16} + \frac{1}{8}

What is the integral from 0 to pi/2 of xsin^2(x) ?
The integral from 0 to pi/2 of xsin^2(x) is (pi^2+2)/(16)+1/8