`int(cosec^2x-2005)/cos^[2005]x.dx`

Options

A. `(cotx)/((cosx)^(2005))+c`
B. `(tan x)/((cosx)^(2005))+c`
C. `(-(tan x))/((cosx)^(2005))+c`
D. none of these

Explanation

To solve the integral \(\int \frac{\csc^2 x - 2005}{\cos^{2005} x} \, dx\), we can break it down step by step.

### Step 1: Rewrite the Integral
We can separate the integral into two parts:
\[ I = \int \frac{\csc^2 x}{\cos^{2005} x} \, dx - 2005 \int \frac{1}{\cos^{2005} x} \, dx \]

### Step 2: Simplify the First Integral
Recall that \(\csc^2 x = \frac{1}{\sin^2 x}\) and \(\sin^2 x = 1 - \cos^2 x\). Thus:
\[ I_1 = \int \frac{\csc^2 x}{\cos^{2005} x} \, dx = \int \frac{1}{\sin^2 x \cos^{2005} x} \, dx \]

### Step 3: Use Integration by Parts
For the first integral, we can use integration by parts. Let:
- \(u = \cos^{-2005} x\) (which is easier to differentiate)
- \(dv = \csc^2 x \, dx\)

Then, we differentiate \(u\) and integrate \(dv\):
- \(du = 2005 \cos^{-2006} x \sin x \, dx\)
- \(v = -\cot x\)

Using integration by parts:
\[ I_1 = u v - \int v \, du \]
Substituting the values:
\[ I_1 = -\cot x \cdot \cos^{-2005} x - \int (-\cot x) \cdot (2005 \cos^{-2006} x \sin x) \, dx \]

### Step 4: Simplify the Integral
Now we can simplify the second integral:
\[ I_1 = -\frac{\cot x}{\cos^{2005} x} + 2005 \int \frac{\sin x}{\cos^{2006} x} \, dx \]

### Step 5: Solve the Second Integral
The second integral can be rewritten as:
\[ I_2 = \int \frac{\sin x}{\cos^{2006} x} \, dx \]
This can be solved by substitution. Let \(u = \cos x\), then \(du = -\sin x \, dx\):
\[ I_2 = -\int u^{-2006} \, du = \frac{u^{-2005}}{2005} + C = \frac{1}{2005 \cos^{2005} x} + C \]

### Step 6: Combine All Parts
Now combining everything back together:
\[ I = -\frac{\cot x}{\cos^{2005} x} + 2005 \left( \frac{1}{2005 \cos^{2005} x} \right) + C \]
The \(2005\) cancels out:
\[ I = -\frac{\cot x}{\cos^{2005} x} + \frac{1}{\cos^{2005} x} + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = -\frac{\cot x}{\cos^{2005} x} + C \]