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

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

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

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

\frac{π ^{2} - 2}{8} - \frac{1}{4}

Decimal

0.73370 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 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}{2}}

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

\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}{2}}

Simplify

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

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

Compute the boundaries:

\frac{π ^{2} - 2}{8} - \frac{1}{4}

= \frac{π ^{2} - 2}{8} - \frac{1}{4}

x \to 0 lim (\frac{( sin ( 4 x ) )}{2 x})

\int 2 sin^{3} (x) cos^{5} (x) d x

\frac{\partial}{\partial y} (x y cos (x y) + sin (x y))

f (x) = arctan (3 x)

\frac{s + 1}{s ^{2} + 4}

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