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

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

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

\int_{0}^{\frac{π}{2}} x^{3} 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}

\frac{d}{d x} (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})^{3}

x \to 2 lim (- x^{2} + 2 x - 6)

\int \frac{1}{a - x ^{2}} d x

x^{3} + 12 x^{2} + 54 x + 14

\int sec^{2} (x) \cdot tan^{5} (x) d x

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