What is the integral from 0 to pi of usin^2(u) ?

The integral from 0 to pi of usin^2(u) is (pi^2)/4

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integral from 0 to pi of usin^2(u)

\int_{0}^{π} u sin^{2} (u) d u

\frac{π ^{2}}{4}

Decimal Notation

2.46740 \dots

Solution steps

\int_{0}^{π} u sin^{2} (u) d u

Apply Integration By Parts

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

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

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

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

Simplify

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

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

(1 + x^{4}) d y + (1 + 4 y^{2}) d x = 0

y^{^{'}} - 5 y^{(2)} e^{(- 3 x)} = 0, y (0) = 9

\int \frac{x ^{3}}{x ^{2} + 2 x + 1} d x

y = \frac{( x ^{2} )}{8 + x}

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