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

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

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

\int_{0}^{π} x^{2} sin^{2} (x) d x

\frac{4 π ^{3} - 6 π}{24}

Decimal Notation

4.38231 \dots

Solution steps

\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}

\frac{sin ( x ^{2} - 5 )}{2 x}

x \to 1 lim (\frac{x ^{2} - 1}{x + 1})

\int u cos (2 u) d u

\frac{df}{d x} f (x) = cos (2 x)

(cos (7 x)) y^{^{'}} - 7 (tan (7 x)) y = 9 sec (7 x)

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