What is the integral of 4r-r^2sin(θ) ?

The integral of 4r-r^2sin(θ) is 2r^2-1/3 r^3sin(θ)+C

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integral of 4r-r^2sin(θ)

\int 4 r - r^{2} sin (θ) d r

2 r^{2} - \frac{1}{3} r^{3} sin (θ) + C

Solution steps

\int 4 r - r^{2} sin (θ) d r

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 r d r - \int sin (θ) r^{2} d r

\int 4 r d r = 2 r^{2}

\int sin (θ) r^{2} d r = \frac{1}{3} r^{3} sin (θ)

= 2 r^{2} - \frac{1}{3} r^{3} sin (θ)

Add a constant to the solution

= 2 r^{2} - \frac{1}{3} r^{3} sin (θ) + C

- 6 x^{2} y + y^{^{'}} = - 10 x^{2}

\frac{1}{s ( s ^{2} + 2 s + 2 )}

\frac{5}{s ^{2} - 12} - \frac{4 s}{s ^{2} + 8}

\frac{\partial}{\partial t} (s t^{2})

y = 3 x^{2} + \frac{8 x}{3} - 2

What is the integral of 4r-r^2sin(θ) ?
The integral of 4r-r^2sin(θ) is 2r^2-1/3 r^3sin(θ)+C