What is the integral of 2sin^3(x/2) ?

The integral of 2sin^3(x/2) is 4(-cos(x/2)+1/3 cos^3(x/2))+C

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integral of 2sin^3(x/2)

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int sin^{3} (\frac{x}{2}) d x

Apply u-substitution

: \int sin^{3} (u) \cdot 2 d u

= 2 \cdot \int sin^{3} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot 2 \cdot \int sin^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int - 1 + v^{2} d v

= 2 \cdot 2 \cdot \int - 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 2 (- \int 1 d v + \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= 2 \cdot 2 (- v + \frac{v ^{3}}{3})

Substitute back

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

Simplify

2 \cdot 2 (- cos (\frac{x}{2}) + \frac{cos ^{3} ( \frac{x}{2} )}{3}) : 4 (- cos (\frac{x}{2}) + \frac{1}{3} cos^{3} (\frac{x}{2}))

= 4 (- cos (\frac{x}{2}) + \frac{1}{3} cos^{3} (\frac{x}{2}))

Add a constant to the solution

= 4 (- cos (\frac{x}{2}) + \frac{1}{3} cos^{3} (\frac{x}{2})) + C

x^{2}, [0, 1]

y^{^{'}} = 6 x + y

\frac{d Q}{d t} = 1 - \frac{Q}{50}

\frac{d}{d x} (\frac{x ^{2}}{x - 2 y})

\frac{6 x ^{2} + 8 x + 4}{x}

What is the integral of 2sin^3(x/2) ?
The integral of 2sin^3(x/2) is 4(-cos(x/2)+1/3 cos^3(x/2))+C