What is the integral from 0 to pi of-6x^2cos(3x) ?

The integral from 0 to pi of-6x^2cos(3x) is (4pi)/3

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integral from 0 to pi of-6x^2cos(3x)

\int_{0}^{π} - 6 x^{2} cos (3 x) d x

\frac{4 π}{3}

Decimal

4.18879 \dots

Solution steps

\int_{0}^{π} - 6 x^{2} cos (3 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 6 \cdot \int_{0}^{π} x^{2} cos (3 x) d x

Apply u-substitution

= - 6 \cdot \int_{0}^{3 π} \frac{u ^{2} cos ( u )}{27} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 6 \cdot \frac{1}{27} \cdot \int_{0}^{3 π} u^{2} cos (u) d u

Apply Integration By Parts

= - 6 \cdot \frac{1}{27} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{3 π}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= - 6 \cdot \frac{1}{27} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{3 π}

Simplify

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

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

Compute the boundaries:

- 6 π

= - \frac{2}{9} (- 6 π)

Simplify

= \frac{4 π}{3}

What is the integral from 0 to pi of-6x^2cos(3x) ?
The integral from 0 to pi of-6x^2cos(3x) is (4pi)/3