What is the integral from 0 to 3pi of 7x^2sin(1/6 x) ?

The integral from 0 to 3pi of 7x^2sin(1/6 x) is 1512(pi-2)

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integral from 0 to 3pi of 7x^2sin(1/6 x)

\int_{0}^{3 π} 7 x^{2} sin (\frac{1}{6} x) d x

1512 (π - 2)

Decimal Notation

1726.08809 \dots

Solution steps

\int_{0}^{3 π} 7 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

= 7 \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

= 7 \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

= 7 \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}}

= 7 \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))

= 7 \cdot 216 [- u^{2} cos (u) - (- 2 (u sin (u) + cos (u)))]_{0}^{\frac{π}{2}}

Simplify

= 1512 [- u^{2} cos (u) + 2 (u sin (u) + cos (u))]_{0}^{\frac{π}{2}}

Compute the boundaries:

π - 2

= 1512 (π - 2)

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\int 15 x 1 - x d x

\int \frac{1}{( x ^{2} + 6 x + 13 )} d x

\frac{d}{d x} (2^{x^{2} + 3 x + 4})

What is the integral from 0 to 3pi of 7x^2sin(1/6 x) ?
The integral from 0 to 3pi of 7x^2sin(1/6 x) is 1512(pi-2)