What is the integral of x^3cos(x/2) ?

The integral of x^3cos(x/2) is 2x^3sin(x/2)-6(-2x^2cos(x/2)+16(1/2 xsin(x/2)+cos(x/2)))+C

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integral of x^3cos(x/2)

\int x^{3} cos (\frac{x}{2}) d x

2 x^{3} sin (\frac{x}{2}) - 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 x^{3} cos (\frac{x}{2}) d x

Apply Integration By Parts

: 2 x^{3} sin (\frac{x}{2}) - \int 6 x^{2} sin (\frac{x}{2}) d x

= 2 x^{3} sin (\frac{x}{2}) - \int 6 x^{2} sin (\frac{x}{2}) d x

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

= 2 x^{3} sin (\frac{x}{2}) - 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

= 2 x^{3} sin (\frac{x}{2}) - 6 (- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))) + C

- 9 x^{2} y + y^{^{'}} = 9 x^{2}

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\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{1}{3}}

\frac{\partial}{\partial y} (4 y^{3} cos (2 x))

\int \frac{1}{x 1 + x} d x

What is the integral of x^3cos(x/2) ?
The integral of x^3cos(x/2) is 2x^3sin(x/2)-6(-2x^2cos(x/2)+16(1/2 xsin(x/2)+cos(x/2)))+C