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

The integral from 0 to pi/4 of x^2cos(2x) is (pi^2-8)/(32)

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

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

\frac{π ^{2} - 8}{32}

Decimal Notation

0.05842 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 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}{4}}

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

\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}{4}}

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

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

Compute the boundaries:

\frac{π ^{2} - 8}{32}

= \frac{π ^{2} - 8}{32}

f (x) = cos (8 x)

y^{^{''}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

\frac{\partial}{\partial x} (cos (2 y - x))

\frac{d}{d x} (lo g_{10} (1 - 2 x^{2}))

(1 + ln (x) + \frac{y}{x}) d x = (6 - ln (x)) d y

What is the integral from 0 to pi/4 of x^2cos(2x) ?
The integral from 0 to pi/4 of x^2cos(2x) is (pi^2-8)/(32)