What is the integral from 0 to pi of 9x^2sin(1/2 x) ?

The integral from 0 to pi of 9x^2sin(1/2 x) is 9(8pi-16)

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integral from 0 to pi of 9x^2sin(1/2 x)

\int_{0}^{π} 9 x^{2} sin (\frac{1}{2} x) d x

9 (8 π - 16)

Decimal

82.19467 \dots

Solution steps

\int_{0}^{π} 9 x^{2} sin (\frac{1}{2} x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{π} x^{2} sin (\frac{1}{2} x) d x

Simplify

\frac{1}{2} x : \frac{x}{2}

= 9 \cdot \int_{0}^{π} x^{2} sin (\frac{x}{2}) d x

Apply Integration By Parts

: [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

= 9 [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

\int - 4 x cos (\frac{x}{2}) d x = - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))

= 9 [- 2 x^{2} cos (\frac{x}{2}) - (- 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2})))]_{0}^{π}

Simplify

= 9 [- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))]_{0}^{π}

Compute the boundaries:

8 π - 16

= 9 (8 π - 16)

\frac{\partial}{\partial x} (- \frac{1}{x ^{2}} + e^{- y})

\frac{d y}{d x} = y (1 + x)^{- 1}

\int \frac{1}{( x + 3 ) ( x + 5 )} d x

\frac{\partial}{\partial x} ((5 x - x^{2}) cos (t))

Y (s) = \frac{s - 2}{s ^{2} - 2 s - 3}

What is the integral from 0 to pi of 9x^2sin(1/2 x) ?
The integral from 0 to pi of 9x^2sin(1/2 x) is 9(8pi-16)