To evaluate the indefinite integral of $\sin^6\left(\frac{x}{3}\right) \cos\left(\frac{x}{3}\right) \, dx$ using substitution, follow these steps:

Step 1: Choose the Substitution

Let: $u = \sin\left(\frac{x}{3}\right)$ Then, differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{1}{3} \cos\left(\frac{x}{3}\right)$ Thus, $du = \frac{1}{3} \cos\left(\frac{x}{3}\right) dx$ or equivalently, $3 \, du = \cos\left(\frac{x}{3}\right) dx$

Step 2: Substitute into the Integral

Substitute $u = \sin\left(\frac{x}{3}\right)$ and $\cos\left(\frac{x}{3}\right) dx = 3 \, du$ into the integral: $\int \sin^6\left(\frac{x}{3}\right) \cos\left(\frac{x}{3}\right) \, dx = \int u^6 \cdot 3 \, du$ This simplifies to: $3 \int u^6 \, du$

Step 3: Integrate

Now, integrate with respect to $u$: $3 \int u^6 \, du = 3 \cdot \frac{u^7}{7} + C$ $= \frac{3}{7} u^7 + C$

Step 4: Substitute Back

Finally, substitute back $u = \sin\left(\frac{x}{3}\right)$ to get the result in terms of $x$: $\frac{3}{7} \sin^7\left(\frac{x}{3}\right) + C$

So, the indefinite integral is: $\int \sin^6\left(\frac{x}{3}\right) \cos\left(\frac{x}{3}\right) \, dx = \frac{3}{7} \sin^7\left(\frac{x}{3}\right) + C$

Would you like further details or have any questions?

Related Questions:

1. How do you evaluate indefinite integrals using substitution for more complex trigonometric expressions?
2. What is the significance of the constant of integration in indefinite integrals?
3. Can you explain the method of substitution for integrals with multiple trigonometric functions?
4. How would the integral change if the exponent on $\sin(x/3)$ were different?
5. What are some common mistakes to avoid when using substitution for integration?

Tip: Always check if the substitution simplifies the integral significantly before proceeding, as it can save time and reduce the risk of errors.