`intsin^4 xcos^3x dx=`

Options

A. `(sin^5x)/(5)+(sin^7x)/(7)+c`
B. `(sin^5x)/(5)-(sin^7x)/(7)+c`
C. `(sin^7x)/(7)+(sin^5x)/(5)+c`
D. `(-sin^7x)/(7)-(sin^5x)/(5)+c`

Answer

To solve the integral \( I = \int \sin^4 x \cos^3 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We can express \( \cos^3 x \) as \( \cos^2 x \cdot \cos x \):
\[ I = \int \sin^4 x \cos^2 x \cos x \, dx \]

### Step 2: Use the Pythagorean identity
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we substitute:
\[ I = \int \sin^4 x (1 - \sin^2 x) \cos x \, dx \]

### Step 3: Substitute \( u = \sin x \)
Let \( u = \sin x \). Then, \( du = \cos x \, dx \). The integral becomes:
\[ I = \int u^4 (1 - u^2) \, du \]

### Step 4: Expand the integrand
Now, expand the integrand:
\[ I = \int (u^4 - u^6) \, du \]

### Step 5: Integrate term by term
Now we can integrate each term separately:
\[ I = \int u^4 \, du - \int u^6 \, du \]
\[ I = \frac{u^5}{5} - \frac{u^7}{7} + C \]

### Step 6: Substitute back \( u = \sin x \)
Now, substituting back \( u = \sin x \):
\[ I = \frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C \]

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