`int tan^(4)x dx = A tan^(3) x+ B tan x + f(x)`, then

Options

A. `A=1/3, B=-1, f(x)=x+C`
B. `A=2/3, B=-1, f(x)= x+C`
C. `A=1/3, B=1, f(x)=x+C`
D. `A=2/3, B=1, f(x)=-x+C`

Answer

To solve the integral \( \int \tan^4 x \, dx \), we can use the identity \( \tan^2 x = \sec^2 x - 1 \). Let's proceed step by step.

### Step 1: Rewrite the Integral
We start by rewriting \( \tan^4 x \):
\[ \tan^4 x = \tan^2 x \cdot \tan^2 x = (\sec^2 x - 1)^2 \]
Thus, we can express the integral as:
\[ \int \tan^4 x \, dx = \int (\sec^2 x - 1)^2 \, dx \]

### Step 2: Expand the Expression
Next, we expand \( (\sec^2 x - 1)^2 \):
\[ (\sec^2 x - 1)^2 = \sec^4 x - 2\sec^2 x + 1 \]
So, we can rewrite the integral:
\[ \int \tan^4 x \, dx = \int (\sec^4 x - 2\sec^2 x + 1) \, dx \]

### Step 3: Split the Integral
Now, we can split the integral into three separate integrals:
\[ \int \tan^4 x \, dx = \int \sec^4 x \, dx - 2\int \sec^2 x \, dx + \int 1 \, dx \]

### Step 4: Solve Each Integral
1. **Integral of \( \sec^4 x \)**:
\[ \int \sec^4 x \, dx = \frac{1}{3} \tan^3 x + C_1 \]
(This can be derived using integration by parts or known results.)

2. **Integral of \( \sec^2 x \)**:
\[ \int \sec^2 x \, dx = \tan x + C_2 \]

3. **Integral of \( 1 \)**:
\[ \int 1 \, dx = x + C_3 \]

### Step 5: Combine the Results
Now, substituting back into our equation:
\[ \int \tan^4 x \, dx = \left( \frac{1}{3} \tan^3 x \right) - 2(\tan x) + x + C \]
where \( C = C_1 - 2C_2 + C_3 \) is a constant of integration.

### Step 6: Final Expression
Thus, we can express the integral as:
\[ \int \tan^4 x \, dx = \frac{1}{3} \tan^3 x - 2 \tan x + x + C \]

### Step 7: Comparing with the Given Form
The problem states that:
\[ \int \tan^4 x \, dx = A \tan^3 x + B \tan x + f(x) \]
From our result:
- \( A = \frac{1}{3} \)
- \( B = -2 \)
- \( f(x) = x + C \)

### Conclusion
Thus, the values are:
- \( A = \frac{1}{3} \)
- \( B = -2 \)
- \( f(x) = x + C \)