What is the integral of 1+2cos(2θ)+1/2 (1+cos(4θ)) ?

The integral of 1+2cos(2θ)+1/2 (1+cos(4θ)) is 3/2 θ+1/8 sin(4θ)+sin(2θ)+C

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integral of 1+2cos(2θ)+1/2 (1+cos(4θ))

\int 1 + 2 cos (2 θ) + \frac{1}{2} (1 + cos (4 θ)) d θ

\frac{3}{2} θ + \frac{1}{8} sin (4 θ) + sin (2 θ) + C

Solution steps

\int 1 + 2 cos (2 θ) + \frac{1}{2} (1 + cos (4 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 1 d θ + \int 2 cos (2 θ) d θ + \int \frac{1}{2} (1 + cos (4 θ)) d θ

\int 1 d θ = θ

\int 2 cos (2 θ) d θ = sin (2 θ)

\int \frac{1}{2} (1 + cos (4 θ)) d θ = \frac{1}{2} (θ + \frac{1}{4} sin (4 θ))

= θ + sin (2 θ) + \frac{1}{2} (θ + \frac{1}{4} sin (4 θ))

Simplify

θ + sin (2 θ) + \frac{1}{2} (θ + \frac{1}{4} sin (4 θ)) : \frac{3}{2} θ + \frac{1}{8} sin (4 θ) + sin (2 θ)

= \frac{3}{2} θ + \frac{1}{8} sin (4 θ) + sin (2 θ)

Add a constant to the solution

= \frac{3}{2} θ + \frac{1}{8} sin (4 θ) + sin (2 θ) + C

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What is the integral of 1+2cos(2θ)+1/2 (1+cos(4θ)) ?
The integral of 1+2cos(2θ)+1/2 (1+cos(4θ)) is 3/2 θ+1/8 sin(4θ)+sin(2θ)+C