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

The integral from 0 to pi/2 of x^2cos(3x) is (-9pi^2+8)/(108)

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

\int_{0}^{\frac{π}{2}} x^{2} cos (3 x) d x

\frac{- 9 π ^{2} + 8}{108}

Decimal Notation

- 0.74839 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x^{2} cos (3 x) d x

Apply u-substitution

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

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

Apply Integration By Parts

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

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

\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}^{\frac{3 π}{2}}

Compute the boundaries:

- \frac{9 π ^{2}}{4} + 2

= \frac{1}{27} (- \frac{9 π ^{2}}{4} + 2)

Simplify

= \frac{- 9 π ^{2} + 8}{108}

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What is the integral from 0 to pi/2 of x^2cos(3x) ?
The integral from 0 to pi/2 of x^2cos(3x) is (-9pi^2+8)/(108)