What is the integral of sqrt(2-2cos(t)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(2-2cos(t))

\int 2 - 2 cos (t) d t

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

Solution steps

\int 2 - 2 cos (t) d t

Apply u-substitution

: \int \frac{4 u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

= \int \frac{4 u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Take the constant out:

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

= 4 \cdot \int \frac{u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Apply u-substitution

: \int \frac{1}{2 v v} d v

= 4 \cdot \int \frac{1}{2 v v} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{v v} d v

Simplify

v v : v^{\frac{3}{2}}

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{3}{2}}} d v

Apply the Power Rule

: - \frac{2}{v}

= 4 \cdot \frac{1}{2} (- \frac{2}{v})

Substitute back

= 4 \cdot \frac{1}{2} - \frac{2}{1 + tan ^{2} ( \frac{t}{2} )}

Simplify

4 \cdot \frac{1}{2} - \frac{2}{1 + tan ^{2} ( \frac{t}{2} )} : - 4 cos (\frac{t}{2})

= - 4 cos (\frac{t}{2})

Add a constant to the solution

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

\int \frac{x ^{2}}{( 1 + x ) ^{3}} d x

\frac{s e ^{- 2 s}}{s ^{2} + 3 s + 2}

\int 7^{4 x} d x

x \to \frac{1}{2} lim (\frac{6 x - 1}{x})

\frac{d}{d x} (arccosh (e^{2 x}))

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