The value of the integral `int("cosec"^(2)x-2019)/(cos^(2019)x)dx` is equal to (where C is the constant of integration)

Options

A. `(cotx)/((cosx)^(2019))+C`
B. `(-cotx)/((cosx)^(2019))+C`
C. `cotx(cosx)^(2019)+C`
D. `-cotx(cosx)^(2019)+C`

Answer

To solve the integral

\[ I = \int \frac{\csc^2 x - 2019}{\cos^{2019} x} \, dx, \]

we can separate the integral into two parts:

\[ I = \int \frac{\csc^2 x}{\cos^{2019} x} \, dx - 2019 \int \frac{1}{\cos^{2019} x} \, dx. \]

### Step 1: Rewrite the first integral
We know that \(\csc^2 x = \frac{1}{\sin^2 x}\), so we can rewrite the first integral as:

\[ I_1 = \int \frac{1}{\sin^2 x \cos^{2019} x} \, dx. \]

### Step 2: Rewrite the second integral
The second integral can be rewritten as:

\[ I_2 = 2019 \int \sec^{2019} x \, dx, \]

where \(\sec x = \frac{1}{\cos x}\).

### Step 3: Use integration by parts
For \(I_1\), we can use integration by parts. Let:

- \(u = \csc^2 x\), then \(du = -2 \csc^2 x \cot x \, dx\).
- \(dv = \frac{1}{\cos^{2019} x} \, dx\).

We can find \(v\) by integrating \(\sec^{2019} x\).

### Step 4: Integrate \(I_1\)
Now we apply integration by parts:

\[ I_1 = \csc^2 x \cdot v - \int v \cdot du. \]

### Step 5: Solve for \(I_2\)
For \(I_2\), we can use the known integral formula for \(\sec^n x\):

\[ \int \sec^n x \, dx = \frac{1}{n-1} \sec^{n-2} x \tan x + \frac{n-2}{n-1} \int \sec^{n-2} x \, dx. \]

### Step 6: Combine results
After calculating both integrals \(I_1\) and \(I_2\), we combine them to find the final result:

\[ I = I_1 - I_2 + C, \]

where \(C\) is the constant of integration.

### Final Result
Thus, the value of the integral is:

\[ I = -\frac{\cot x}{\cos^{2019} x} + \frac{2019}{2018} \sec^{2019} x + C. \]