What is the integral from 0 to (3pi)/4 of x^3sin(2x) ?

The integral from 0 to (3pi)/4 of x^3sin(2x) is (3(-9pi^2+8))/(64)

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integral from 0 to (3pi)/4 of x^3sin(2x)

\int_{0}^{\frac{3 π}{4}} x^{3} sin (2 x) d x

\frac{3 ( - 9 π ^{2} + 8 )}{64}

Decimal Notation

- 3.78873 \dots

Solution steps

\int_{0}^{\frac{3 π}{4}} x^{3} sin (2 x) d x

Apply Integration By Parts

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

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

\int - \frac{3}{2} x^{2} cos (2 x) d x = - \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= [\frac{1}{2} (- x^{3} cos (2 x) - 2 (- \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{0}^{\frac{3 π}{4}}

Simplify

[\frac{1}{2} (- x^{3} cos (2 x) - 2 (- \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{0}^{\frac{3 π}{4}} : [\frac{1}{2} (- x^{3} cos (2 x) + \frac{3}{4} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)))]_{0}^{\frac{3 π}{4}}

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

Compute the boundaries:

\frac{3 ( - 9 π ^{2} + 8 )}{64}

= \frac{3 ( - 9 π ^{2} + 8 )}{64}

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What is the integral from 0 to (3pi)/4 of x^3sin(2x) ?
The integral from 0 to (3pi)/4 of x^3sin(2x) is (3(-9pi^2+8))/(64)