`int"cosec"^(4)x dx ` is equal to

Options

A. `cotx+(cot^(3))/(3)+C`
B. `tanx+(tan^(3)x)/(3)+C`
C. `-cotx-(cot^(3)x)/(3)+C`
D. `-tanx-(tan^(3)x)/(3)+C`

Explanation

To solve the integral \( \int \csc^4 x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \csc^4 x \, dx \]
We can express \( \csc^4 x \) as \( \csc^2 x \cdot \csc^2 x \):
\[ \int \csc^4 x \, dx = \int \csc^2 x \cdot \csc^2 x \, dx \]

### Step 2: Use the Identity
We know that \( \csc^2 x = 1 + \cot^2 x \). Therefore, we can rewrite the integral:
\[ \int \csc^4 x \, dx = \int (1 + \cot^2 x) \csc^2 x \, dx \]

### Step 3: Distribute the Integral
Now we can separate the integral:
\[ \int \csc^4 x \, dx = \int \csc^2 x \, dx + \int \cot^2 x \csc^2 x \, dx \]

### Step 4: Integrate the First Part
The integral of \( \csc^2 x \) is:
\[ \int \csc^2 x \, dx = -\cot x \]

### Step 5: Integrate the Second Part
For the second integral \( \int \cot^2 x \csc^2 x \, dx \), we can use the substitution \( u = \cot x \), which gives \( du = -\csc^2 x \, dx \). Thus, we have:
\[ \int \cot^2 x \csc^2 x \, dx = -\int u^2 \, du \]
Integrating \( u^2 \):
\[ -\int u^2 \, du = -\frac{u^3}{3} = -\frac{\cot^3 x}{3} \]

### Step 6: Combine the Results
Now we combine both parts of the integral:
\[ \int \csc^4 x \, dx = -\cot x - \frac{\cot^3 x}{3} + C \]

### Final Answer
Thus, the final result is:
\[ \int \csc^4 x \, dx = -\cot x - \frac{\cot^3 x}{3} + C \]

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