What is the integral of 6x^2sin(1/2 x) ?

The integral of 6x^2sin(1/2 x) is 6(-2x^2cos(x/2)+16(1/2 xsin(x/2)+cos(x/2)))+C

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integral of 6x^2sin(1/2 x)

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

6 (- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))) + C

Solution steps

\int 6 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

= 6 \cdot \int x^{2} sin (\frac{1}{2} x) d x

Simplify

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

= 6 \cdot \int 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

= 6 (- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x)

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

= 6 (- 2 x^{2} cos (\frac{x}{2}) - (- 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))))

Simplify

= 6 (- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2})))

Add a constant to the solution

= 6 (- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))) + C

\frac{x ^{2} - 1}{x + 3}

e^{2 - t}

(t - 4)^{2}

\frac{\partial}{\partial y} (x^{3})

(x^{4} + y^{4}) d x - x y^{3} d y = 0

What is the integral of 6x^2sin(1/2 x) ?
The integral of 6x^2sin(1/2 x) is 6(-2x^2cos(x/2)+16(1/2 xsin(x/2)+cos(x/2)))+C