If `int(sin^(4)x)/(cos^(8)x)dx=atan^(7)x +btan^(5)x+C` , then

Options

A. 7a = 5b
B. 5a = 7b
C. 7a + 5b = 0
D. 5a + 7b = 0

Explanation

To solve the integral \( \int \frac{\sin^4 x}{\cos^8 x} \, dx \) and find the constants \( a \) and \( b \) in the expression \( \int \frac{\sin^4 x}{\cos^8 x} \, dx = a \tan^7 x + b \tan^5 x + C \), we will follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ \int \frac{\sin^4 x}{\cos^8 x} \, dx = \int \frac{\sin^4 x}{\cos^4 x \cos^4 x} \, dx = \int \tan^4 x \sec^4 x \, dx \]

### Step 2: Use Substitution
Let \( t = \tan x \). Then, the derivative \( dt = \sec^2 x \, dx \) implies \( dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \).

### Step 3: Substitute in the Integral
Now, substituting \( \tan x = t \) into the integral:
\[ \int \tan^4 x \sec^4 x \, dx = \int t^4 \sec^4 x \cdot \frac{dt}{1 + t^2} \]
Using the identity \( \sec^2 x = 1 + \tan^2 x = 1 + t^2 \), we have \( \sec^4 x = (1 + t^2)^2 \).
Thus,
\[ \int t^4 (1 + t^2)^2 \cdot \frac{dt}{1 + t^2} = \int t^4 (1 + t^2) \, dt \]

### Step 4: Expand and Integrate
Now, we expand:
\[ \int t^4 (1 + t^2) \, dt = \int (t^4 + t^6) \, dt \]
Integrating term by term:
\[ = \frac{t^5}{5} + \frac{t^7}{7} + C \]

### Step 5: Substitute Back
Substituting back \( t = \tan x \):
\[ = \frac{\tan^5 x}{5} + \frac{\tan^7 x}{7} + C \]

### Step 6: Identify Coefficients
From the expression \( \int \frac{\sin^4 x}{\cos^8 x} \, dx = a \tan^7 x + b \tan^5 x + C \), we can identify:
- Coefficient of \( \tan^7 x \) is \( a = \frac{1}{7} \)
- Coefficient of \( \tan^5 x \) is \( b = \frac{1}{5} \)

### Step 7: Establish Relationships
Now we need to find the relationships between \( a \) and \( b \) based on the options provided:
1. \( 7a = 5b \)
2. \( 5a = 7b \)
3. \( 7a + 5b = 0 \)
4. \( 5a + 7b = 0 \)

Substituting the values:
- For \( 7a = 5b \):
\[ 7 \cdot \frac{1}{7} = 5 \cdot \frac{1}{5} \implies 1 = 1 \quad \text{(True)} \]
- For \( 5a = 7b \):
\[ 5 \cdot \frac{1}{7} = 7 \cdot \frac{1}{5} \implies \frac{5}{7} \neq \frac{7}{5} \quad \text{(False)} \]
- For \( 7a + 5b = 0 \):
\[ 7 \cdot \frac{1}{7} + 5 \cdot \frac{1}{5} = 1 + 1 \neq 0 \quad \text{(False)} \]
- For \( 5a + 7b = 0 \):
\[ 5 \cdot \frac{1}{7} + 7 \cdot \frac{1}{5} = \frac{5}{7} + \frac{7}{5} \neq 0 \quad \text{(False)} \]

### Conclusion
The correct relationship is \( 7a = 5b \).