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

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

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

\int_{0}^{π} (x sin (x))^{2} d x

\frac{4 π ^{3} - 6 π}{24}

Decimal

4.38231 \dots

Solution steps

\int_{0}^{π} (x sin (x))^{2} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= \int_{0}^{π} x^{2} sin^{2} (x) d x

Apply Integration By Parts

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

= [\frac{1}{3} (x^{3} sin^{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} sin^{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} sin^{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} sin^{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} sin^{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}

\int_{0}^{2} (x - 3) e^{- x} d x

\int_{0}^{2} - 4 x^{4} + 8 x^{3} d x

\int_{0}^{4} (4 x^{3} + \frac{2}{x + 1}) d x

\int_{- 2}^{0} \frac{1}{x ^{2} + 2 x + 2} d x

\int_{- 1}^{0} \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

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