Evaluate the following integrals:
`intsin^(7)(3-2x)dx`

Answer

To evaluate the integral \( \int \sin^7(3 - 2x) \, dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( u = 3 - 2x \). Then, we differentiate to find \( du \):
\[ du = -2 \, dx \quad \Rightarrow \quad dx = -\frac{1}{2} du \]

### Step 2: Rewrite the Integral
Substituting \( u \) and \( dx \) into the integral gives:
\[ \int \sin^7(3 - 2x) \, dx = \int \sin^7(u) \left(-\frac{1}{2}\right) du = -\frac{1}{2} \int \sin^7(u) \, du \]

### Step 3: Break Down \( \sin^7(u) \)
We can express \( \sin^7(u) \) as:
\[ \sin^7(u) = \sin^6(u) \sin(u) = \sin^6(u)(1 - \cos^2(u)) \]
This can be expanded as:
\[ \sin^7(u) = \sin^6(u) - \sin^6(u) \cos^2(u) \]

### Step 4: Substitute and Split the Integral
Now, we can rewrite the integral:
\[ -\frac{1}{2} \int \sin^7(u) \, du = -\frac{1}{2} \left( \int \sin^6(u) \, du - \int \sin^6(u) \cos^2(u) \, du \right) \]

### Step 5: Use the Identity for \( \sin^6(u) \)
Using the identity \( \sin^2(u) = 1 - \cos^2(u) \):
\[ \sin^6(u) = (1 - \cos^2(u))^3 \]
Now we can expand this:
\[ \sin^6(u) = 1 - 3\cos^2(u) + 3\cos^4(u) - \cos^6(u) \]

### Step 6: Integrate Each Term
Now we need to integrate each term separately:
\[ \int (1 - 3\cos^2(u) + 3\cos^4(u) - \cos^6(u)) \, du \]
This requires using the substitution \( t = \cos(u) \) and \( dt = -\sin(u) \, du \).

### Step 7: Final Integration
After integrating each term and substituting back for \( u \):
\[ \int \sin^7(u) \, du = \text{(integrated terms)} + C \]
Substituting back \( u = 3 - 2x \) will give us the final answer.

### Final Answer
The final expression will be:
\[ -\frac{1}{2} \left( \text{integrated terms in terms of } u \right) + C \]