`int(1)/(sinx*cos^(2)x)dx=`

Options

A. `secx+log|secx+tanx|+C`
B. `secx*tanx+C`
C. `secx+log|secx-tanx|+C`
D. `secx+log|"cosec x"-cotx|+C`

Explanation

To solve the integral \( \int \frac{1}{\sin x \cos^2 x} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
Let \( I = \int \frac{1}{\sin x \cos^2 x} \, dx \).

### Step 2: Use the Pythagorean Identity
We know that \( \sin^2 x + \cos^2 x = 1 \). To simplify the integral, we can express it as:
\[ I = \int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^2 x} \, dx \]

### Step 3: Split the Integral
Now, we can separate the integral into two parts:
\[ I = \int \left( \frac{\sin^2 x}{\sin x \cos^2 x} + \frac{\cos^2 x}{\sin x \cos^2 x} \right) \, dx \]
This simplifies to:
\[ I = \int \left( \frac{\sin x}{\cos^2 x} + \frac{1}{\sin x} \right) \, dx \]

### Step 4: Rewrite the Terms
We can rewrite the first term:
\[ I = \int \tan x \sec x \, dx + \int \csc x \, dx \]

### Step 5: Integrate Each Term
1. The integral of \( \tan x \sec x \) is \( \sec x + C_1 \).
2. The integral of \( \csc x \) is \( -\ln |\csc x + \cot x| + C_2 \).

Combining these results, we have:
\[ I = \sec x - \ln |\csc x + \cot x| + C \]
where \( C = C_1 + C_2 \) is the constant of integration.

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{1}{\sin x \cos^2 x} \, dx = \sec x - \ln |\csc x + \cot x| + C \]

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