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 Notation

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}}

= [\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}

\frac{d}{d x} (cos (x) + ln (cos (x)) + x sin (x))

\int \frac{x ^{3}}{1 + x ^{3}} d x

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

f (x) = 4 e^{x} - x^{16} x, at x = 0

\frac{\partial}{\partial x} (π \cdot (\frac{x}{2})^{2} h)

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