`int1/(cosx(1+cosx))dx=`

Options

A. `logabs(secx+tanx)+2tan""x/2+c`
B. `logabs(secx+tanx)-2tan""x/2+c`
C. `logabs(secx+tanx)+tan""x/2+c`
D. `logabs(secx+tanx)-tan""x/2+c`

Answer

To solve the integral \( \int \frac{1}{\cos x (1 + \cos x)} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We can rewrite the integrand as:
\[ \frac{1}{\cos x (1 + \cos x)} = \frac{1}{\cos x} \cdot \frac{1}{1 + \cos x} \]
This allows us to separate the integral:
\[ \int \frac{1}{\cos x (1 + \cos x)} \, dx = \int \frac{1}{\cos x} \cdot \frac{1}{1 + \cos x} \, dx \]

### Step 2: Use the identity for \(1 + \cos x\)
We know that:
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the integral:
\[ \int \frac{1}{\cos x (1 + \cos x)} \, dx = \int \frac{1}{\cos x \cdot 2 \cos^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \frac{1}{\cos x \cos^2\left(\frac{x}{2}\right)} \, dx \]

### Step 3: Use the identity for \(\cos x\)
We also know that:
\[ \cos x = 2 \cos^2\left(\frac{x}{2}\right) - 1 \]
Thus:
\[ \frac{1}{\cos x} = \frac{1}{2 \cos^2\left(\frac{x}{2}\right) - 1} \]
Now we can substitute this back into our integral:
\[ \frac{1}{2} \int \frac{1}{(2 \cos^2\left(\frac{x}{2}\right) - 1) \cos^2\left(\frac{x}{2}\right)} \, dx \]

### Step 4: Substitute \(u = \tan\left(\frac{x}{2}\right)\)
We can use the Weierstrass substitution:
\[ \cos x = \frac{1 - u^2}{1 + u^2}, \quad dx = \frac{2}{1 + u^2} \, du \]
Substituting these into the integral gives:
\[ \frac{1}{2} \int \frac{1 + u^2}{(1 - u^2)(1 + u^2)^2} \cdot \frac{2}{1 + u^2} \, du = \int \frac{1}{(1 - u^2)(1 + u^2)} \, du \]

### Step 5: Partial Fraction Decomposition
Now we can perform partial fraction decomposition:
\[ \frac{1}{(1 - u^2)(1 + u^2)} = \frac{A}{1 - u^2} + \frac{B}{1 + u^2} \]
Solving for \(A\) and \(B\) gives:
\[ 1 = A(1 + u^2) + B(1 - u^2) \]
Setting \(u = 1\) and \(u = -1\) allows us to solve for \(A\) and \(B\).

### Step 6: Integrate
After finding \(A\) and \(B\), we can integrate each term separately:
\[ \int \frac{A}{1 - u^2} \, du + \int \frac{B}{1 + u^2} \, du \]

### Step 7: Back Substitute
Finally, we will back substitute \(u = \tan\left(\frac{x}{2}\right)\) to express our answer in terms of \(x\).

### Final Answer
The final result will be:
\[ \int \frac{1}{\cos x (1 + \cos x)} \, dx = \text{(result from integration)} + C \]

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