If `intcos^(7) xdx =Asin^(7)x+Bsin^(5)x+C sin^(3)x+sinx+k`, then

Options

A. `A=(1)/(7),B=(3)/(5),C=-1`
B. `A=-(1)/(7),B=(3)/(5),C=-1`
C. `A=(-1)/(7),B=(1)/(5),C=-1`
D. `A=(1)/(7),B=(3)/(5),C=1`

Explanation

To solve the integral \( \int \cos^7 x \, dx \) and express it in the form \( A \sin^7 x + B \sin^5 x + C \sin^3 x + \sin x + k \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \cos^7 x \, dx \]
We can express \( \cos^7 x \) as:
\[ \int \cos x \cdot \cos^6 x \, dx \]

### Step 2: Use the Pythagorean Identity
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we rewrite \( \cos^6 x \):
\[ \cos^6 x = (\cos^2 x)^3 = (1 - \sin^2 x)^3 \]
Thus, we have:
\[ \int \cos x (1 - \sin^2 x)^3 \, dx \]

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

### Step 4: Expand the Expression
Now we expand \( (1 - t^2)^3 \):
\[ (1 - t^2)^3 = 1 - 3t^2 + 3t^4 - t^6 \]
So, the integral now is:
\[ \int (1 - 3t^2 + 3t^4 - t^6) \, dt \]

### Step 5: Integrate Each Term
Now we integrate term by term:
\[ \int 1 \, dt = t \]
\[ \int -3t^2 \, dt = -t^3 \]
\[ \int 3t^4 \, dt = \frac{3}{5} t^5 \]
\[ \int -t^6 \, dt = -\frac{1}{7} t^7 \]
Putting it all together, we have:
\[ t - t^3 + \frac{3}{5} t^5 - \frac{1}{7} t^7 + C \]

### Step 6: Substitute Back \( t = \sin x \)
Now we substitute back \( t = \sin x \):
\[ \sin x - \sin^3 x + \frac{3}{5} \sin^5 x - \frac{1}{7} \sin^7 x + C \]

### Step 7: Rearranging the Terms
Rearranging the terms gives us:
\[ -\frac{1}{7} \sin^7 x + \frac{3}{5} \sin^5 x - \sin^3 x + \sin x + C \]

### Step 8: Identify Coefficients
Now we can identify the coefficients:
- \( A = -\frac{1}{7} \)
- \( B = \frac{3}{5} \)
- \( C = -1 \)

### Final Answer
Thus, the values of \( A \), \( B \), and \( C \) are:
\[ A = -\frac{1}{7}, \quad B = \frac{3}{5}, \quad C = -1 \]

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