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

The integral from 0 to pi of x^3sin(3x) is (9pi^3-6pi)/(27)

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

\int_{0}^{π} x^{3} sin (3 x) d x

\frac{9 π ^{3} - 6 π}{27}

Decimal Notation

9.63729 \dots

Solution steps

\int_{0}^{π} x^{3} sin (3 x) d x

Apply Integration By Parts

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

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

\int - x^{2} cos (3 x) d x = - \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

= [\frac{1}{3} (- x^{3} cos (3 x) - 3 (- \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))))]_{0}^{π}

Simplify

[\frac{1}{3} (- x^{3} cos (3 x) - 3 (- \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))))]_{0}^{π} : [\frac{1}{27} (- 9 x^{3} cos (3 x) + 9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))]_{0}^{π}

= [\frac{1}{27} (- 9 x^{3} cos (3 x) + 9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))]_{0}^{π}

Compute the boundaries:

\frac{9 π ^{3} - 6 π}{27}

= \frac{9 π ^{3} - 6 π}{27}

\int_{1}^{2} (x + 2 y) d x

\int_{1}^{4} (- 3 x) d x

\int_{0}^{4} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

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

\int_{- 2}^{1} x^{2} + x + 2 d x

What is the integral from 0 to pi of x^3sin(3x) ?
The integral from 0 to pi of x^3sin(3x) is (9pi^3-6pi)/(27)