What is the integral from 0 to pi/4 of x^2sin(2x) ?

The integral from 0 to pi/4 of x^2sin(2x) is pi/8-1/4

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integral from 0 to pi/4 of x^2sin(2x)

\int_{0}^{\frac{π}{4}} x^{2} sin (2 x) d x

\frac{π}{8} - \frac{1}{4}

Decimal

0.14269 \dots

Solution steps

\int_{0}^{\frac{π}{4}} x^{2} sin (2 x) d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π}{8} - \frac{1}{4}

= \frac{π}{8} - \frac{1}{4}

What is the integral from 0 to pi/4 of x^2sin(2x) ?
The integral from 0 to pi/4 of x^2sin(2x) is pi/8-1/4