It looks like you're working with an integral involving trigonometric functions. Let me carefully interpret and solve the integral you described:

The integral seems to be:

$\int \sin(2\theta) \sqrt{\cos(2\theta)} \left(1 - \cos^2(2\theta)\right) \, d\theta.$

We will proceed step by step.

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Step 1: Simplify the expression $1 - \cos^2(2\theta)$

This expression can be rewritten using a Pythagorean identity:

$1 - \cos^2(2\theta) = \sin^2(2\theta).$

So, the integral becomes:

$\int \sin(2\theta) \sqrt{\cos(2\theta)} \cdot \sin^2(2\theta) \, d\theta.$

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Step 2: Rearrange the integrand

Now we rewrite the integral to group terms:

$\int \sin^3(2\theta) \cdot \sqrt{\cos(2\theta)} \, d\theta.$

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Step 3: Substitution

Let's use the substitution $u = \cos(2\theta)$. Then:

$\frac{du}{d\theta} = -2 \sin(2\theta) \quad \Rightarrow \quad d\theta = -\frac{du}{2\sin(2\theta)}.$

Also, notice that:

$\sin^2(2\theta) = 1 - u^2.$

Thus, the integral becomes:

$-\frac{1}{2} \int (1 - u^2) \cdot \sqrt{u} \, du.$

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Step 4: Expand and Simplify

Now distribute the $\sqrt{u}$:

$-\frac{1}{2} \int \left(\sqrt{u} - u^{5/2}\right) \, du.$

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Step 5: Integrate each term

We now integrate both terms separately.

1. $\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$,
2. $\int u^{5/2} \, du = \frac{2}{7} u^{7/2}$.

Thus, the integral becomes:

$-\frac{1}{2} \left( \frac{2}{3} u^{3/2} - \frac{2}{7} u^{7/2} \right) + C.$

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Step 6: Substitute $u = \cos(2\theta)$

Now, substitute back $u = \cos(2\theta)$:

$-\frac{1}{2} \left( \frac{2}{3} \cos^{3/2}(2\theta) - \frac{2}{7} \cos^{7/2}(2\theta) \right) + C.$

Simplify:

$-\frac{1}{3} \cos^{3/2}(2\theta) + \frac{1}{7} \cos^{7/2}(2\theta) + C.$

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Final Answer

$\int \sin(2\theta) \sqrt{\cos(2\theta)} \left(1 - \cos^2(2\theta)\right) \, d\theta = -\frac{1}{3} \cos^{3/2}(2\theta) + \frac{1}{7} \cos^{7/2}(2\theta) + C.$

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Would you like me to go over any step in detail or provide further clarifications?

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Here are 5 related questions for further exploration:

1. How do you perform substitution in integrals involving trigonometric functions?
2. What are some common trigonometric identities used in calculus?
3. How do you integrate powers of trigonometric functions?
4. What is the difference between definite and indefinite integrals?
5. How do you approach integrals involving products of trigonometric functions?

Tip: Substitution is one of the most useful techniques for integrating complex trigonometric functions—look for patterns involving derivatives of inner functions.