What is the integral of rrθ*2-4sin(θ) ?

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

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integral of rrθ*2-4sin(θ)

integral

rr θ \cdot 2 - 4 sin (θ)

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

Solution steps

\int rr θ \cdot 2 - 4 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 sin (θ) d r + \int 2 θ rr d r

\int 4 sin (θ) d r = 4 sin (θ) r

\int 2 θ rr d r = \frac{2 θ r ^{3}}{3}

= - 4 sin (θ) r + \frac{2 θ r ^{3}}{3}

Add a constant to the solution

= - 4 sin (θ) r + \frac{2 θ r ^{3}}{3} + C

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