What is the integral of (1-cos(t/2))^2sin(t/2) ?

The integral of (1-cos(t/2))^2sin(t/2) is 2/3 (1-cos(t/2))^3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-cos(t/2))^2sin(t/2)

\int (1 - cos (\frac{t}{2}))^{2} sin (\frac{t}{2}) d t

\frac{2}{3} (1 - cos (\frac{t}{2}))^{3} + C

Solution steps

\int (1 - cos (\frac{t}{2}))^{2} sin (\frac{t}{2}) d t

Apply u-substitution

: \int 2 u^{2} d u

= \int 2 u^{2} d u

Take the constant out:

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

= 2 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 2 \cdot \frac{u ^{3}}{3}

Substitute back

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

= 2 \cdot \frac{( 1 - cos ( \frac{t}{2} ) ) ^{3}}{3}

Simplify

2 \cdot \frac{( 1 - cos ( \frac{t}{2} ) ) ^{3}}{3} : \frac{2}{3} (1 - cos (\frac{t}{2}))^{3}

= \frac{2}{3} (1 - cos (\frac{t}{2}))^{3}

Add a constant to the solution

= \frac{2}{3} (1 - cos (\frac{t}{2}))^{3} + C

\int 2 x e^{2 x^{2}} d x

\frac{d y}{d x} = y - x^{2}

y^{^{''}} - 10 y^{^{'}} + 25 = 0

\frac{d}{d x} ((\frac{cos ( x )}{1 + sin ( x )})^{5})

\int \frac{x ^{3} + 2 x ^{2} + 3 x - 2}{x - 2} d x

What is the integral of (1-cos(t/2))^2sin(t/2) ?
The integral of (1-cos(t/2))^2sin(t/2) is 2/3 (1-cos(t/2))^3+C