The given problem involves an integral where we are asked to evaluate it using the substitution $u = 1 - \sin \left(\frac{t}{4}\right)$.

The integral is:

$\int \left(1 - \sin \left( \frac{t}{4} \right) \right)^2 \cdot \cos \left( \frac{t}{4} \right) \, dt$

with the substitution $u = 1 - \sin \left( \frac{t}{4} \right)$.

Step 1: Differentiate the substitution

First, we differentiate $u = 1 - \sin \left( \frac{t}{4} \right)$ to find $du$:

$\frac{du}{dt} = -\frac{1}{4} \cos \left( \frac{t}{4} \right)$

or equivalently:

$du = -\frac{1}{4} \cos \left( \frac{t}{4} \right) \, dt$

Multiplying both sides by $-4$ gives:

$-4 \, du = \cos \left( \frac{t}{4} \right) \, dt$

Step 2: Rewrite the integral in terms of $u$

Using the substitution $u = 1 - \sin \left( \frac{t}{4} \right)$, the integral becomes:

$\int u^2 \cdot (-4 \, du)$

This simplifies to:

$-4 \int u^2 \, du$

Step 3: Solve the integral

The integral of $u^2$ is:

$\int u^2 \, du = \frac{u^3}{3}$

Thus, the integral becomes:

$-4 \cdot \frac{u^3}{3} = -\frac{4}{3} u^3$

Step 4: Substitute back $u$

Now, substitute $u = 1 - \sin \left( \frac{t}{4} \right)$ back into the expression:

$-\frac{4}{3} \left(1 - \sin \left( \frac{t}{4} \right) \right)^3 + C$

Step 5: Compare with answer choices

This matches the expression in option C:

$-\frac{4}{3} \left(1 - \sin \left( \frac{t}{4} \right)\right)^3 + C$

Conclusion

The correct answer is:

C.

Would you like further explanation or any details? Here are some additional questions to expand on this problem:

1. How does substitution simplify solving integrals?
2. Why is the factor $-4$ important when using substitution?
3. How would you approach solving an integral like this without substitution?
4. What are the advantages of recognizing patterns in integrals for quick solving?
5. How does one differentiate complex trigonometric functions?

Tip: Always verify your substitution by differentiating the result to ensure the transformation is correct.