What is the integral of 8sin^2(x)cos(x) ?

The integral of 8sin^2(x)cos(x) is 8/3 sin^3(x)+C

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integral of 8sin^2(x)cos(x)

\int 8 sin^{2} (x) cos (x) d x

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

Solution steps

\int 8 sin^{2} (x) cos (x) d x

Take the constant out:

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

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

Apply u-substitution

: \int u^{2} d u

= 8 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 8 \cdot \frac{u ^{3}}{3}

Substitute back

u = sin (x)

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

Simplify

8 \cdot \frac{sin ^{3} ( x )}{3} : \frac{8}{3} sin^{3} (x)

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

Add a constant to the solution

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

y^{^{''}} + 2 y^{^{'}} + y = 3 x + 8

\frac{d}{d x} (a cos (2 x) + b cos (2 x))

\frac{\partial}{\partial x} (y - 5 x)

\int x [2 cos (x) - x sin (x)] d x

x \to 2 lim (6 - 3 x - 3)

What is the integral of 8sin^2(x)cos(x) ?
The integral of 8sin^2(x)cos(x) is 8/3 sin^3(x)+C