`int(cos^7x)/(sinx)dx=`

Options

A. `log|sinx|-(3)/(2)sin^2x+(3)/(4)sin^4x-(1)/(6)sin^6x+c`
B. `log|sinx|-(3)/(4)sin^2x+(3)/(2)sin^4x-(1)/(6)sin^6x+c`
C. `log|sinx|-(3)/(2)sin^2x+(3)/(2)sin^4x-(1)/(6)sin^6x+c`
D. `log|sinx|-(3)/(4)sin^2x+(3)/(4)sin^4x-(1)/(6)sin^6x+c`

Answer

To solve the integral \( I = \int \frac{\cos^7 x}{\sin x} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int \frac{\cos^7 x}{\sin x} \, dx = \int \frac{\cos^6 x \cdot \cos x}{\sin x} \, dx \]

### Step 2: Use the Identity for Cosine
We can express \( \cos^6 x \) in terms of \( \sin x \):
\[ \cos^6 x = \cos^6 x = (1 - \sin^2 x)^3 \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{(1 - \sin^2 x)^3 \cos x}{\sin x} \, dx \]

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

### Step 4: Expand the Integrand
Now, we expand \( (1 - t^2)^3 \):
\[ (1 - t^2)^3 = 1 - 3t^2 + 3t^4 - t^6 \]
Thus, we have:
\[ I = \int \frac{1 - 3t^2 + 3t^4 - t^6}{t} \, dt = \int \left( \frac{1}{t} - 3t + 3t^3 - t^5 \right) \, dt \]

### Step 5: Integrate Term by Term
Now we can integrate each term:
\[ I = \int \frac{1}{t} \, dt - 3 \int t \, dt + 3 \int t^3 \, dt - \int t^5 \, dt \]
Calculating each integral:
1. \( \int \frac{1}{t} \, dt = \ln |t| \)
2. \( \int t \, dt = \frac{t^2}{2} \)
3. \( \int t^3 \, dt = \frac{t^4}{4} \)
4. \( \int t^5 \, dt = \frac{t^6}{6} \)

Putting it all together:
\[ I = \ln |t| - \frac{3t^2}{2} + \frac{3t^4}{4} - \frac{t^6}{6} + C \]

### Step 6: Substitute Back
Now we substitute back \( t = \sin x \):
\[ I = \ln |\sin x| - \frac{3 \sin^2 x}{2} + \frac{3 \sin^4 x}{4} - \frac{\sin^6 x}{6} + C \]

### Final Answer
Thus, the final answer is:
\[ I = \ln |\sin x| - \frac{\sin^6 x}{6} - \frac{3 \sin^2 x}{2} + \frac{3 \sin^4 x}{4} + C \]