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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 + lim (15 cos (15 x))

\int \frac{2 x ^{3} + x ^{2} - x - 31}{x ^{2} - x - 6} d x

\frac{d}{d x} (\frac{x}{x ^{2} + 2 x + 2})

3 x^{2} - 12 x - 15

\frac{\partial}{\partial x} (e^{- 2 t} cos (x))

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