What is the integral from 0 to 3pi of 6x^2sin(1/6 x) ?

The integral from 0 to 3pi of 6x^2sin(1/6 x) is 1296(pi-2)

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integral from 0 to 3pi of 6x^2sin(1/6 x)

\int_{0}^{3 π} 6 x^{2} sin (\frac{1}{6} x) d x

1296 (π - 2)

Decimal

1479.50407 \dots

Solution steps

\int_{0}^{3 π} 6 x^{2} sin (\frac{1}{6} 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}^{3 π} x^{2} sin (\frac{1}{6} x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} 216 u^{2} sin (u) d u

= 6 \cdot \int_{0}^{\frac{π}{2}} 216 u^{2} sin (u) d u

Take the constant out:

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

= 6 \cdot 216 \cdot \int_{0}^{\frac{π}{2}} u^{2} sin (u) d u

Apply Integration By Parts

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

= 6 \cdot 216 [- u^{2} cos (u) - \int - 2 u cos (u) d u]_{0}^{\frac{π}{2}}

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

= 6 \cdot 216 [- u^{2} cos (u) - (- 2 (u sin (u) + cos (u)))]_{0}^{\frac{π}{2}}

Simplify

= 1296 [- u^{2} cos (u) + 2 (u sin (u) + cos (u))]_{0}^{\frac{π}{2}}

Compute the boundaries:

π - 2

= 1296 (π - 2)

\int_{0}^{6} - x^{2} + 6 x d x

20 x^{3}

\int_{1}^{6} 5 \frac{ln ( x )}{x ^{2}} d x

f (x) = cos (2 x)

f (x) = (x^{3} - 4)^{2} (2 x + 5)^{3}

What is the integral from 0 to 3pi of 6x^2sin(1/6 x) ?
The integral from 0 to 3pi of 6x^2sin(1/6 x) is 1296(pi-2)