What is the integral of (1-cos(2x))^2 ?

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

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integral of (1-cos(2x))^2

\int (1 - cos (2 x))^{2} d x

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

Solution steps

\int (1 - cos (2 x))^{2} d x

Apply u-substitution

: \int \frac{1}{2} - cos (u) + \frac{1}{2} cos^{2} (u) d u

= \int \frac{1}{2} - cos (u) + \frac{1}{2} cos^{2} (u) d u

Apply the Sum Rule:

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

= \int \frac{1}{2} d u - \int cos (u) d u + \int \frac{1}{2} cos^{2} (u) d u

\int \frac{1}{2} d u = \frac{1}{2} u

\int cos (u) d u = sin (u)

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

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

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} \cdot 2 x - sin (2 x) + \frac{1}{4} (2 x + \frac{1}{2} sin (2 \cdot 2 x)) : \frac{3}{2} x + \frac{1}{8} sin (4 x) - sin (2 x)

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

Add a constant to the solution

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

\frac{d ^{3}}{d x ^{3}} (\frac{1}{x})

\int (5 x^{2} + 6 + \frac{7}{x ^{2} + 1}) d x

\frac{d}{d x} (- cosh (x))

f (x) = \frac{9}{x ^{2}}, at x = a

n = 2 \sum \infty \frac{1}{n \cdot ( n + 1 )}

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