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

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

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integral of-cos^3(x)

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

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

Solution steps

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

Take the constant out:

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

= - \int cos^{3} (x) d x

Rewrite using trig identities

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

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 (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (sin (3) x^{2} cos (2 x))

\int t \cdot e^{t} d t

y^{^{''}} - y^{^{'}} + 121 y = 11 sin (11 t)

\int \frac{1}{x x - 2} d x

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{2 x}

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