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

The integral from 0 to pi of xsin^2(x) is (pi^2)/4

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

\int_{0}^{π} x sin^{2} (x) d x

\frac{π ^{2}}{4}

Decimal

2.46740 \dots

Solution steps

\int_{0}^{π} 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}^{π}

= [\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}^{π}

\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}^{π}

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

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

Compute the boundaries:

\frac{2 π ^{2} - 1}{8} + \frac{1}{8}

= \frac{2 π ^{2} - 1}{8} + \frac{1}{8}

Simplify

= \frac{π ^{2}}{4}

14 x - 3 x^{2}

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

\int (4 x - x^{2}) d x

\frac{\partial}{\partial x} (- 2 sin (2 x))

arcsec (3 x)

What is the integral from 0 to pi of xsin^2(x) ?
The integral from 0 to pi of xsin^2(x) is (pi^2)/4