What is the integral of sqrt(1-cos(4x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(1-cos(4x))

\int 1 - cos (4 x) d x

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

Solution steps

\int 1 - cos (4 x) d x

Apply Integral Substitution

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

Rewrite using trig identities

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

Take the constant out:

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

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

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

assuming

sin (u) \geq 0

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (1 + x x)

\frac{5}{2 x ^{6}}

x \to - 2 lim ((x + 1)^{7} (3 x^{2}))

\int - \frac{x}{y} d x

y^{^{'}} ln (y) + x^{2} y = 0

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