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

The integral of-cos^3(t) is -sin(t)+1/3 sin^3(t)+C

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integral of-cos^3(t)

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

- sin (t) + \frac{1}{3} sin^{3} (t) + C

Solution steps

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

Take the constant out:

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

= - \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})

Simplify

- (sin (t) - \frac{sin ^{3} ( t )}{3}) : - sin (t) + \frac{1}{3} sin^{3} (t)

= - sin (t) + \frac{1}{3} sin^{3} (t)

Add a constant to the solution

= - sin (t) + \frac{1}{3} sin^{3} (t) + C

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