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

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

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integral of 2sin^3(u)

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

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

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

\int 1 d v = v

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

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

Substitute back

v = cos (u)

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

Add a constant to the solution

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

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