What is the integral of 4+8cos(θ)+4cos^2(θ) ?

The integral of 4+8cos(θ)+4cos^2(θ) is 6θ+8sin(θ)+sin(2θ)+C

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integral of 4+8cos(θ)+4cos^2(θ)

\int 4 + 8 cos (θ) + 4 cos^{2} (θ) d θ

6 θ + 8 sin (θ) + sin (2 θ) + C

Solution steps

\int 4 + 8 cos (θ) + 4 cos^{2} (θ) d θ

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 d θ + \int 8 cos (θ) d θ + \int 4 cos^{2} (θ) d θ

\int 4 d θ = 4 θ

\int 8 cos (θ) d θ = 8 sin (θ)

\int 4 cos^{2} (θ) d θ = 2 θ + sin (2 θ)

= 4 θ + 8 sin (θ) + 2 θ + sin (2 θ)

Simplify

= 6 θ + 8 sin (θ) + sin (2 θ)

Add a constant to the solution

= 6 θ + 8 sin (θ) + sin (2 θ) + C

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What is the integral of 4+8cos(θ)+4cos^2(θ) ?
The integral of 4+8cos(θ)+4cos^2(θ) is 6θ+8sin(θ)+sin(2θ)+C