What is the integral of sqrt(1-cos(8x)) ?

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

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integral of sqrt(1-cos(8x))

\int 1 - cos (8 x) d x

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

Solution steps

\int 1 - cos (8 x) d x

Apply Integral Substitution

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

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 u) d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{2} 1 - cos (2 v) d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 - cos (2 v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 2 sin^{2} (v) d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} 2 \cdot \int sin^{2} (v) d v

sin^{2} (v) = (sin (v)),

assuming

sin (v) \geq 0

= \frac{1}{2} \cdot \frac{1}{2} 2 \cdot \int sin (v) d v

Use the common integral:

\int sin (v) d v = - cos (v)

= \frac{1}{2} \cdot \frac{1}{2} 2 (- cos (v))

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} 2 (- cos (2 \cdot 2 x))

Simplify

\frac{1}{2} \cdot \frac{1}{2} 2 (- cos (2 \cdot 2 x)) : - \frac{1}{2 2} cos (4 x)

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

Add a constant to the solution

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

x \to 3 lim (\frac{1 - x}{x - 3})

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\frac{d}{d x} (x^{n} (ln (x))^{2})

f (x) = \frac{9 - x e ^{x}}{x + e ^{x}}

\frac{d}{d x} ((5 x - 2)^{3})

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