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

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

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integral of (1/4 (1-cos^2(4x)))

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

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

Solution steps

\int \frac{1}{4} (1 - cos^{2} (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 \int 1 - 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

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