What is the integral of (4-cos(3x))^2 ?

The integral of (4-cos(3x))^2 is 33/2 x-8/3 sin(3x)+1/12 sin(6x)+C

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integral of (4-cos(3x))^2

\int (4 - cos (3 x))^{2} d x

\frac{33}{2} x - \frac{8}{3} sin (3 x) + \frac{1}{12} sin (6 x) + C

Solution steps

\int (4 - cos (3 x))^{2} d x

Factor

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

= \int (- (- 4 + cos (3 x)))^{2} d x

Simplify

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

= \int (- 4 + cos (3 x))^{2} d x

Expand

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

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

\int 16 d x = 16 x

\int 8 cos (3 x) d x = \frac{8}{3} sin (3 x)

\int cos^{2} (3 x) d x = \frac{1}{2} (x + \frac{1}{6} sin (6 x))

= 16 x - \frac{8}{3} sin (3 x) + \frac{1}{2} (x + \frac{1}{6} sin (6 x))

Simplify

16 x - \frac{8}{3} sin (3 x) + \frac{1}{2} (x + \frac{1}{6} sin (6 x)) : \frac{33}{2} x - \frac{8}{3} sin (3 x) + \frac{1}{12} sin (6 x)

= \frac{33}{2} x - \frac{8}{3} sin (3 x) + \frac{1}{12} sin (6 x)

Add a constant to the solution

= \frac{33}{2} x - \frac{8}{3} sin (3 x) + \frac{1}{12} sin (6 x) + C

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What is the integral of (4-cos(3x))^2 ?
The integral of (4-cos(3x))^2 is 33/2 x-8/3 sin(3x)+1/12 sin(6x)+C