What is the integral from 0 to 1 of x^2sin^2(x) ?

The integral from 0 to 1 of x^2sin^2(x) is 1/3 sin^2(1)-1/8 sin(2)-1/12 cos(2)

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integral from 0 to 1 of x^2sin^2(x)

\int_{0}^{1} x^{2} sin^{2} (x) d x

\frac{1}{3} sin^{2} (1) - \frac{1}{8} sin (2) - \frac{1}{12} cos (2)

Decimal Notation

0.15704 \dots

Solution steps

\int_{0}^{1} x^{2} sin^{2} (x) d x

Apply Integration By Parts

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

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

\int \frac{1}{3} x^{3} sin (2 x) 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) - (- \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}^{1}

Expand

\frac{1}{3} x^{3} sin^{2} (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) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x)

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

Compute the boundaries:

\frac{1}{3} sin^{2} (1) - \frac{1}{8} sin (2) - \frac{1}{12} cos (2)

= \frac{1}{3} sin^{2} (1) - \frac{1}{8} sin (2) - \frac{1}{12} cos (2)

\int_{1}^{8} (1 + x^{3}) d x

\int_{0}^{\frac{π}{2}} sin (x) + cos (x) d x

\int_{0}^{2} 4 x - x^{2} d x

\int_{- π}^{π} sin (x) sin (2 x) d x

\int_{5}^{5} 4 x d x

What is the integral from 0 to 1 of x^2sin^2(x) ?
The integral from 0 to 1 of x^2sin^2(x) is 1/3 sin^2(1)-1/8 sin(2)-1/12 cos(2)