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

The integral of (2-cos(x))^2 is 9/2 x+1/4 sin(2x)-4sin(x)+C

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

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

\frac{9}{2} x + \frac{1}{4} sin (2 x) - 4 sin (x) + C

Solution steps

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

Simplify

(2 - cos (x))^{2} : (2 - cos (x)) (2 - cos (x))

= \int (2 - cos (x)) (2 - cos (x)) d x

Expand

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

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

\int 4 d x = 4 x

\int 4 cos (x) d x = 4 sin (x)

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

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

Simplify

4 x - 4 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{9}{2} x + \frac{1}{4} sin (2 x) - 4 sin (x)

= \frac{9}{2} x + \frac{1}{4} sin (2 x) - 4 sin (x)

Add a constant to the solution

= \frac{9}{2} x + \frac{1}{4} sin (2 x) - 4 sin (x) + C

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\int 12 x d x

\int \frac{- x}{1 + x ^{2}} d x

What is the integral of (2-cos(x))^2 ?
The integral of (2-cos(x))^2 is 9/2 x+1/4 sin(2x)-4sin(x)+C