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

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

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integral of 1-cos^2(2x)

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (4 x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 - cos (4 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

= \frac{1}{2} (\int 1 d x - \int cos (4 x) d x)

\int 1 d x = x

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

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

Add a constant to the solution

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

R (x) = 71 x

\int \frac{3}{5 y y ^{2} - 1} d y

\frac{d}{d x} (- \frac{1 - x}{2})

8 e^{x} + x

x \to - 1 lim (\frac{x}{1 - x ^{4}})

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