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

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

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integral of-sin^3(x)

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

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

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{4}{cos ( x ) - 1} d x

y^{^{''}} + 49 y = sin (7 t)

x \to 1 lim (\frac{x - 8 + 7}{x - 1})

y = \frac{x}{2 x + \frac{- 4}{x}}

9 y y^{^{'}} + 4 x = 0

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