`int(1)/(tanx+cotx)dx=`

Options

A. `log(secx+cosx)+c`
B. `2cos 2x+c`
C. `-(1)/(2)cos 2x+c`
D. `(1)/(2)sin^(2)x+c`

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting \( \tan x \) and \( \cot x \) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \]
Thus, we have:
\[ \tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \]

### Step 2: Find a common denominator
To combine the fractions, we find a common denominator:
\[ \tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can simplify this to:
\[ \tan x + \cot x = \frac{1}{\sin x \cos x} \]

### Step 3: Substitute into the integral
Now we substitute this back into the integral:
\[ \int \frac{1}{\tan x + \cot x} \, dx = \int \sin x \cos x \, dx \]

### Step 4: Use a trigonometric identity
We can use the double angle identity for sine:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]
Thus, the integral becomes:
\[ \int \sin x \cos x \, dx = \int \frac{1}{2} \sin 2x \, dx \]

### Step 5: Integrate
Now we can integrate:
\[ \int \frac{1}{2} \sin 2x \, dx = \frac{1}{2} \cdot \left(-\frac{1}{2} \cos 2x\right) + C = -\frac{1}{4} \cos 2x + C \]

### Final answer
Therefore, the solution to the integral is:
\[ \int \frac{1}{\tan x + \cot x} \, dx = -\frac{1}{4} \cos 2x + C \]

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