What is the integral from 0 to 2pi of (4-cos(3x))^2 ?

The integral from 0 to 2pi of (4-cos(3x))^2 is 33pi

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integral from 0 to 2pi of (4-cos(3x))^2

\int_{0}^{2 π} (4 - cos (3 x))^{2} d x

33 π

Solution steps

\int_{0}^{2 π} (4 - cos (3 x))^{2} d x

Factor

4 - cos (3 x) : - (- 4 + cos (3 x))

= \int_{0}^{2 π} (- (- 4 + cos (3 x)))^{2} d x

Simplify

(- (- 4 + cos (3 x)))^{2} : (- 4 + cos (3 x))^{2}

= \int_{0}^{2 π} (- 4 + cos (3 x))^{2} d x

Expand

(- 4 + cos (3 x))^{2} : 16 - 8 cos (3 x) + cos^{2} (3 x)

= \int_{0}^{2 π} 16 - 8 cos (3 x) + cos^{2} (3 x) d x

Apply the Sum Rule:

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

= \int_{0}^{2 π} 16 d x - \int_{0}^{2 π} 8 cos (3 x) d x + \int_{0}^{2 π} cos^{2} (3 x) d x

\int_{0}^{2 π} 16 d x = 32 π

\int_{0}^{2 π} 8 cos (3 x) d x = 0

\int_{0}^{2 π} cos^{2} (3 x) d x = π

= 32 π - 0 + π

Simplify

= 33 π

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What is the integral from 0 to 2pi of (4-cos(3x))^2 ?
The integral from 0 to 2pi of (4-cos(3x))^2 is 33pi