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

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

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integral of cos^3(t)

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

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

Solution steps

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

Rewrite using trig identities

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

= sin (t) - \frac{sin ^{3} ( t )}{3}

Add a constant to the solution

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

- y^{^{'}} + \frac{3 y}{x} = x^{2} y^{2}

\int_{1}^{9} \frac{2}{x} d x

\frac{d}{d x} (- 3 csc (x^{\frac{2}{3}}))

\int x^{- 1} + sin (x) + e^{x} d x

\frac{d y}{d x} + 2 x y = 1

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