`int(cosx)/(sin^(2)x)dx=`

Options

A. `secx`
B. `-cotx`
C. `tanx`
D. `cotx`

Answer

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

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form:
\[ \int \frac{\cos x}{\sin^2 x} \, dx = \int \cos x \cdot \frac{1}{\sin^2 x} \, dx \]

### Step 2: Recognize Trigonometric Identities
We know that:
\[ \frac{\cos x}{\sin x} = \cot x \quad \text{and} \quad \frac{1}{\sin x} = \csc x \]
Thus, we can rewrite the integral as:
\[ \int \cos x \cdot \csc^2 x \, dx \]

### Step 3: Use Substitution
Now, we can use the substitution:
\[ u = \sin x \quad \Rightarrow \quad du = \cos x \, dx \]
This means that \(dx = \frac{du}{\cos x}\). Substituting in gives us:
\[ \int \frac{\cos x}{\sin^2 x} \, dx = \int \frac{1}{u^2} \, du \]

### Step 4: Integrate
Now we can integrate:
\[ \int u^{-2} \, du = -\frac{1}{u} + C \]
Substituting back \(u = \sin x\):
\[ -\frac{1}{\sin x} + C = -\csc x + C \]

### Final Answer
Thus, the final answer is:
\[ -\csc x + C \]

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