To find the area bounded between the curves \(y = 6 \sin x\) and \(y + 8 \sin^3 x = 0\) from \(x = 0\) to \(x = \pi\), we will follow these steps:
### Step 1: Identify the curves
The first curve is given by:
\[
y_1 = 6 \sin x
\]
The second curve can be rearranged as:
\[
y_2 = -8 \sin^3 x
\]
### Step 2: Set up the area integral
The area \(A\) between the two curves from \(x = 0\) to \(x = \pi\) can be expressed as:
\[
A = \int_{0}^{\pi} (y_1 - y_2) \, dx
\]
Substituting the expressions for \(y_1\) and \(y_2\):
\[
A = \int_{0}^{\pi} (6 \sin x - (-8 \sin^3 x)) \, dx
\]
This simplifies to:
\[
A = \int_{0}^{\pi} (6 \sin x + 8 \sin^3 x) \, dx
\]
### Step 3: Break down the integral
We can split the integral into two parts:
\[
A = \int_{0}^{\pi} 6 \sin x \, dx + \int_{0}^{\pi} 8 \sin^3 x \, dx
\]
### Step 4: Calculate the first integral
The first integral is straightforward:
\[
\int 6 \sin x \, dx = -6 \cos x
\]
Evaluating from \(0\) to \(\pi\):
\[
\left[-6 \cos x\right]_{0}^{\pi} = -6 \cos(\pi) - (-6 \cos(0)) = -6(-1) - (-6(1)) = 6 + 6 = 12
\]
### Step 5: Calculate the second integral
For the second integral, we can use the identity \(\sin^3 x = \sin x (1 - \cos^2 x)\):
\[
\int 8 \sin^3 x \, dx = 8 \int \sin x (1 - \cos^2 x) \, dx = 8 \left( \int \sin x \, dx - \int \sin x \cos^2 x \, dx \right)
\]
The first part is:
\[
\int \sin x \, dx = -\cos x
\]
Evaluating from \(0\) to \(\pi\):
\[
\left[-\cos x\right]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2
\]
Thus, the first part contributes \(8 \cdot 2 = 16\).
For the second part, we can use substitution:
Let \(u = \cos x\), then \(du = -\sin x \, dx\). The limits change from \(x = 0\) (where \(u = 1\)) to \(x = \pi\) (where \(u = -1\)):
\[
\int \sin x \cos^2 x \, dx = -\int_{1}^{-1} u^2 \, du = \int_{-1}^{1} u^2 \, du
\]
Calculating this integral:
\[
\int u^2 \, du = \frac{u^3}{3}
\]
Evaluating from \(-1\) to \(1\):
\[
\left[\frac{u^3}{3}\right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}
\]
Thus, the second part contributes:
\[
8 \cdot \frac{2}{3} = \frac{16}{3}
\]
### Step 6: Combine the results
Now, we can combine the two parts:
\[
A = 12 + \frac{16}{3} = \frac{36}{3} + \frac{16}{3} = \frac{52}{3}
\]
### Final Answer
Thus, the area bounded between the curves from \(x = 0\) to \(x = \pi\) is:
\[
\boxed{\frac{68}{3}}
\]
