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

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

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integral of sin^3(u)

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

= \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

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

\int 1 d v = v

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

= - v + \frac{v ^{3}}{3}

Substitute back

v = cos (u)

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

Add a constant to the solution

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

\frac{d}{d x} ((ln (x))^{- 1})

y = x^{\frac{4}{3}}

\frac{1}{s ^{2} - 1}

\int_{0}^{π} cos (x) sin (x) d x

\int_{0}^{\frac{π}{6}} 64 sin^{2} (x) d x

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