What is the integral of 8cos^3(x) ?

The integral of 8cos^3(x) is 8(sin(x)-(sin^3(x))/3)+C

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integral of 8cos^3(x)

\int 8 cos^{3} (x) d x

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

Solution steps

\int 8 cos^{3} (x) d x

Take the constant out:

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

= 8 \cdot \int cos^{3} (x) d x

Rewrite using trig identities

= 8 \cdot \int (1 - sin^{2} (x)) cos (x) d x

Apply u-substitution

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

= 8 \cdot \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

= 8 (\int 1 d u - \int u^{2} d u)

\int 1 d u = u

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

= 8 (u - \frac{u ^{3}}{3})

Substitute back

u = sin (x)

= 8 (sin (x) - \frac{sin ^{3} ( x )}{3})

Add a constant to the solution

= 8 (sin (x) - \frac{sin ^{3} ( x )}{3}) + C

\frac{\partial}{\partial x} (p (x))

\frac{d}{d x} (x^{2} sin^{4} (x) + x cos^{- 2} (x))

y^{^{'}} - 6 y = 2 e^{t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = - 4, y^{^{'}} (0) = - 6

x \to + 2 lim ([7 f (x) + g (x)])

What is the integral of 8cos^3(x) ?
The integral of 8cos^3(x) is 8(sin(x)-(sin^3(x))/3)+C