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

The integral from 0 to pi of t^2cos(3t) is -(2pi)/9

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

\int_{0}^{π} t^{2} cos (3 t) d t

- \frac{2 π}{9}

Decimal Notation

- 0.69813 \dots

Solution steps

\int_{0}^{π} t^{2} cos (3 t) d t

Apply u-substitution

: \int_{0}^{3 π} \frac{u ^{2} cos ( u )}{27} d u

= \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

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

Apply Integration By Parts

: [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{3 π}

= \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))

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

Compute the boundaries:

- 6 π

= \frac{1}{27} (- 6 π)

Simplify

= - \frac{2 π}{9}

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What is the integral from 0 to pi of t^2cos(3t) ?
The integral from 0 to pi of t^2cos(3t) is -(2pi)/9