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

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

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integral of sin^3(t)

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

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 u + \int u^{2} d u

\int 1 d u = u

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

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

Substitute back

u = cos (t)

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

Add a constant to the solution

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

11 (t + 1) \frac{d y}{d t} - 9 y = 18 t

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

\int 5 x e^{- 3 x} d x

\frac{d y}{d x} = x + 9

x \to 2 - lim (\frac{x}{x ^{2} - 4})

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