What is the integral from 0 to 1 of xsin^2(x) ?

The integral from 0 to 1 of xsin^2(x) is 3/8-1/4 sin(2)-1/8 cos(2)

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integral from 0 to 1 of xsin^2(x)

\int_{0}^{1} x sin^{2} (x) d x

\frac{3}{8} - \frac{1}{4} sin (2) - \frac{1}{8} cos (2)

Decimal

0.19969 \dots

Solution steps

\int_{0}^{1} x sin^{2} (x) d x

Apply Integration By Parts

: [\frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x]_{0}^{1}

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

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

Simplify

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

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

Compute the boundaries:

\frac{3}{8} - \frac{1}{4} sin (2) - \frac{1}{8} cos (2)

= \frac{3}{8} - \frac{1}{4} sin (2) - \frac{1}{8} cos (2)

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What is the integral from 0 to 1 of xsin^2(x) ?
The integral from 0 to 1 of xsin^2(x) is 3/8-1/4 sin(2)-1/8 cos(2)