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

The integral from 0 to pi of (xcos(x))^2 is (4pi^3+6pi)/(24)

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

\int_{0}^{π} (x cos (x))^{2} d x

\frac{4 π ^{3} + 6 π}{24}

Decimal Notation

5.95311 \dots

Solution steps

\int_{0}^{π} (x cos (x))^{2} d x

Simplify

(x cos (x))^{2} : x^{2} cos^{2} (x)

= \int_{0}^{π} x^{2} cos^{2} (x) d x

Apply Integration By Parts

: [\frac{1}{3} (x^{3} cos^{2} (x) - 3 \cdot \int - \frac{x ^{3} sin ( 2 x )}{3} d x)]_{0}^{π}

= [\frac{1}{3} (x^{3} cos^{2} (x) - 3 \cdot \int - \frac{x ^{3} sin ( 2 x )}{3} d x)]_{0}^{π}

\int - \frac{x ^{3} sin ( 2 x )}{3} d x = \frac{1}{6} x^{3} cos (2 x) - \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= [\frac{1}{3} (x^{3} cos^{2} (x) - 3 (\frac{1}{6} x^{3} cos (2 x) - \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{0}^{π}

Simplify

[\frac{1}{3} (x^{3} cos^{2} (x) - 3 (\frac{1}{6} x^{3} cos (2 x) - \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{0}^{π} : [\frac{1}{24} (8 x^{3} cos^{2} (x) - 4 x^{3} cos (2 x) + 6 x^{2} sin (2 x) + 6 x cos (2 x) - 3 sin (2 x))]_{0}^{π}

= [\frac{1}{24} (8 x^{3} cos^{2} (x) - 4 x^{3} cos (2 x) + 6 x^{2} sin (2 x) + 6 x cos (2 x) - 3 sin (2 x))]_{0}^{π}

Compute the boundaries:

\frac{4 π ^{3} + 6 π}{24}

= \frac{4 π ^{3} + 6 π}{24}

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What is the integral from 0 to pi of (xcos(x))^2 ?
The integral from 0 to pi of (xcos(x))^2 is (4pi^3+6pi)/(24)