(i) `int tan dx=log |secx|+c`
(ii) `int cot x dx=log |sin x|+c`
(iii) `int tan x dx=log |sin x| +c`
(iv) `int tan x dx =-log cos x +c`

Answer

To solve the given integrals, we will evaluate each of the four statements one by one.

### (i) \( \int \tan x \, dx = \log |\sec x| + C \)

1. **Rewrite the integral**:
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus,
\[ \int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx \]

2. **Use substitution**:
Let \( u = \cos x \). Then, \( du = -\sin x \, dx \) or \( \sin x \, dx = -du \).
So, the integral becomes:
\[ \int \frac{\sin x}{\cos x} \, dx = -\int \frac{1}{u} \, du \]

3. **Integrate**:
\[ -\int \frac{1}{u} \, du = -\log |u| + C = -\log |\cos x| + C \]

4. **Rewrite the result**:
Using the property of logarithms:
\[ -\log |\cos x| = \log |\sec x| + C \]
Thus,
\[ \int \tan x \, dx = \log |\sec x| + C \]

**Conclusion**: The first statement is **true**.

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### (ii) \( \int \cot x \, dx = \log |\sin x| + C \)

1. **Rewrite the integral**:
\[ \cot x = \frac{\cos x}{\sin x} \]
Thus,
\[ \int \cot x \, dx = \int \frac{\cos x}{\sin x} \, dx \]

2. **Use substitution**:
Let \( u = \sin x \). Then, \( du = \cos x \, dx \).
So, the integral becomes:
\[ \int \frac{\cos x}{\sin x} \, dx = \int \frac{1}{u} \, du \]

3. **Integrate**:
\[ \int \frac{1}{u} \, du = \log |u| + C = \log |\sin x| + C \]

**Conclusion**: The second statement is **true**.

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### (iii) \( \int \tan x \, dx = \log |\sin x| + C \)

From our previous calculation, we found:
\[ \int \tan x \, dx = \log |\sec x| + C \]
This does not equal \( \log |\sin x| + C \).

**Conclusion**: The third statement is **false**.

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### (iv) \( \int \tan x \, dx = -\log \cos x + C \)

From our previous calculation, we found:
\[ \int \tan x \, dx = -\log |\cos x| + C \]
This can be rewritten as:
\[ -\log |\cos x| = \log |\sec x| + C \]
Thus, this statement is equivalent to the first statement.

**Conclusion**: The fourth statement is **true**.

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### Summary of Results:
- (i) True
- (ii) True
- (iii) False
- (iv) True