What is the integral from 0 to pi/2 of x^3*cos(2x) ?

The integral from 0 to pi/2 of x^3*cos(2x) is -(3(-2+pi^2))/(16)+3/8

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integral from 0 to pi/2 of x^3*cos(2x)

\int_{0}^{\frac{π}{2}} x^{3} \cdot cos (2 x) d x

- \frac{3 ( - 2 + π ^{2} )}{16} + \frac{3}{8}

Decimal Notation

- 1.10055 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x^{3} cos (2 x) d x

Apply Integration By Parts

: [\frac{1}{2} (x^{3} sin (2 x) - 2 \cdot \int \frac{3}{2} x^{2} sin (2 x) d x)]_{0}^{\frac{π}{2}}

= [\frac{1}{2} (x^{3} sin (2 x) - 2 \cdot \int \frac{3}{2} x^{2} sin (2 x) d x)]_{0}^{\frac{π}{2}}

\int \frac{3}{2} x^{2} sin (2 x) d x = \frac{3}{2} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x)))

= [\frac{1}{2} (x^{3} sin (2 x) - 2 \cdot \frac{3}{2} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{\frac{π}{2}}

Simplify

[\frac{1}{2} (x^{3} sin (2 x) - 2 \cdot \frac{3}{2} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{\frac{π}{2}} : [\frac{1}{2} (x^{3} sin (2 x) - 3 (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{\frac{π}{2}}

= [\frac{1}{2} (x^{3} sin (2 x) - 3 (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{\frac{π}{2}}

Compute the boundaries:

- \frac{3 ( - 2 + π ^{2} )}{16} + \frac{3}{8}

= - \frac{3 ( - 2 + π ^{2} )}{16} + \frac{3}{8}

\int_{1}^{4} 3 x^{2} + x x + 1 d x

\int_{- 1}^{1} \frac{1}{4} d x

\int_{1}^{5} sin (x^{2}) d x

\int_{0}^{1} 1 + cos (π x) d x

\int_{0}^{π} x^{2} sin (k x) d x

What is the integral from 0 to pi/2 of x^3*cos(2x) ?
The integral from 0 to pi/2 of x^3*cos(2x) is -(3(-2+pi^2))/(16)+3/8