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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{1}{2} (1 + cos (8 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 (8 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 (8 x) d x)

\int 1 d x = x

\int cos (8 x) d x = \frac{1}{8} sin (8 x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{s} \cdot (\frac{10}{s ^{2} + s + 10})

- 4 e^{3 t}

x \to 0 lim (- 9 x^{2} + 6 x)

csc (2 x)

f (t) = 4 t - 9 t^{2}

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