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

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

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

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

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

Solution steps

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

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{x}{2})

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

Simplify

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

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

Add a constant to the solution

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

\int_{- 1}^{2} (4 - x^{2}) d x

y = lo g_{5} (2) x^{2}

n = 1 \sum \infty 7^{n + 1} x^{n}

\frac{1}{t}

\frac{6}{x ^{2} + 2}, at 2, 1

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