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

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

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integral from 0 to pi of x*sin^2(x)

\int_{0}^{π} x \cdot sin^{2} (x) d x

\frac{π ^{2}}{4}

Decimal Notation

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}

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

\int \frac{( 3 x ^{2} )}{2} d x

\frac{d}{d x} (e^{x^{2} + 6 x})

\int (5 x + 1)^{4} d x

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{2} + 1})

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