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

The integral of (4-cos(x/4))^2 is 33/2 x-32sin(x/4)+sin(x/2)+C

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

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

\frac{33}{2} x - 32 sin (\frac{x}{4}) + sin (\frac{x}{2}) + C

Solution steps

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

Expand

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

= \int 16 - 8 cos (\frac{x}{4}) + cos^{2} (\frac{x}{4}) 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 (\frac{x}{4}) d x + \int cos^{2} (\frac{x}{4}) d x

\int 16 d x = 16 x

\int 8 cos (\frac{x}{4}) d x = 32 sin (\frac{x}{4})

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

= 16 x - 32 sin (\frac{x}{4}) + \frac{1}{2} x + sin (\frac{x}{2})

Simplify

16 x - 32 sin (\frac{x}{4}) + \frac{1}{2} x + sin (\frac{x}{2}) : \frac{33}{2} x - 32 sin (\frac{x}{4}) + sin (\frac{x}{2})

= \frac{33}{2} x - 32 sin (\frac{x}{4}) + sin (\frac{x}{2})

Add a constant to the solution

= \frac{33}{2} x - 32 sin (\frac{x}{4}) + sin (\frac{x}{2}) + C

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What is the integral of (4-cos(x/4))^2 ?
The integral of (4-cos(x/4))^2 is 33/2 x-32sin(x/4)+sin(x/2)+C