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

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

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

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

- \frac{π}{4}

Decimal

- 0.78539 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- \frac{π}{4}

= - \frac{π}{4}

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