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

The integral from 0 to pi of 3xcos(3x) is -2/3

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integral from 0 to pi of 3xcos(3x)

\int_{0}^{π} 3 x cos (3 x) d x

- \frac{2}{3}

Decimal Notation

- 0.66666 \dots

Solution steps

\int_{0}^{π} 3 x cos (3 x) d x

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{3 π} \frac{u cos ( u )}{9} d u

= 3 \cdot \int_{0}^{3 π} \frac{u cos ( u )}{9} d u

Take the constant out:

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

= 3 \cdot \frac{1}{9} \cdot \int_{0}^{3 π} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{3 π}

= 3 \cdot \frac{1}{9} [u sin (u) - \int sin (u) d u]_{0}^{3 π}

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

= 3 \cdot \frac{1}{9} [u sin (u) - (- cos (u))]_{0}^{3 π}

Simplify

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

= \frac{1}{3} [u sin (u) + cos (u)]_{0}^{3 π}

Compute the boundaries:

- 2

= \frac{1}{3} (- 2)

Simplify

= - \frac{2}{3}

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

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\int_{0}^{2 π} cos^{2} (3 x) d x

\int_{0}^{π} x^{3} sin (n x) d x

\int_{0}^{0.4} ln (1 + x^{2}) d x

What is the integral from 0 to pi of 3xcos(3x) ?
The integral from 0 to pi of 3xcos(3x) is -2/3