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

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

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

\int_{0}^{π} x^{2} cos (3 x) d x

- \frac{2 π}{9}

Decimal

- 0.69813 \dots

Solution steps

\int_{0}^{π} x^{2} cos (3 x) d x

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}

\int_{0}^{π} cos^{4} (θ) d θ

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

\int_{0}^{1} 3 x^{2} arcsin (x) d x

\int_{- 5}^{2} \frac{5}{x ^{4}} d x

\int_{0}^{2} - \frac{3}{2} x + 3 d x

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