If `intsecx cosecx dx=log|g(x)|+c`, then what is g (x) equal to ?

Options

A. sinx cosx
B. `sec^(2)x`
C. `tanx`
D. `log|tan|x`

Answer

To solve the integral \( \int \sec x \csc x \, dx \) and find \( g(x) \) such that \( \int \sec x \csc x \, dx = \log |g(x)| + C \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ \int \sec x \csc x \, dx = \int \frac{1}{\cos x \sin x} \, dx \]

2. **Use Trigonometric Identity**:
We can use the identity \( \sin 2x = 2 \sin x \cos x \) to rewrite the integral:
\[ \int \frac{1}{\sin x \cos x} \, dx = \int \frac{2}{\sin 2x} \, dx \]

3. **Factor Out the Constant**:
Factor out the constant 2:
\[ = 2 \int \frac{1}{\sin 2x} \, dx \]

4. **Use the Integral of Cosecant**:
The integral of \( \frac{1}{\sin u} \) is \( \ln |\csc u - \cot u| + C \). Thus, we have:
\[ = 2 \ln |\csc 2x - \cot 2x| + C \]

5. **Express in Terms of \( g(x) \)**:
We need to express this in the form \( \log |g(x)| + C \). Therefore, we can set:
\[ g(x) = \csc 2x - \cot 2x \]

6. **Final Result**:
Thus, we find that:
\[ g(x) = \csc 2x - \cot 2x \]