If `int(dx)/(cos^(6)x+sin^(6)x)=tan^(-1)(-Kcot2x)+C` then

Options

A. K=2
B. k=4
C. k=6
D. k depends on x

Answer

To solve the integral

\[ I = \int \frac{dx}{\cos^6 x + \sin^6 x} \]

and find the value of \( k \) in the equation

\[ I = \tan^{-1}(-K \cot 2x) + C, \]

we will go through the following steps:

### Step 1: Simplify the Denominator
We start by rewriting the denominator \( \cos^6 x + \sin^6 x \). We can use the identity for the sum of cubes:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

Let \( a = \cos^2 x \) and \( b = \sin^2 x \). Then we have:

\[ \cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x) \]

Since \( \cos^2 x + \sin^2 x = 1 \), we simplify to:

\[ \cos^6 x + \sin^6 x = \cos^4 x - \cos^2 x \sin^2 x + \sin^4 x \]

### Step 2: Express in Terms of \( \tan x \)
Next, we can rewrite \( \cos^4 x \) and \( \sin^4 x \):

\[ \cos^4 x = \frac{1}{(1 + \tan^2 x)^2}, \quad \sin^4 x = \frac{\tan^4 x}{(1 + \tan^2 x)^2} \]

Thus, we have:

\[ \cos^6 x + \sin^6 x = \frac{1}{(1 + \tan^2 x)^2} - \frac{\tan^2 x}{(1 + \tan^2 x)^2} + \frac{\tan^4 x}{(1 + \tan^2 x)^2} \]

This simplifies to:

\[ \cos^6 x + \sin^6 x = \frac{1 - \tan^2 x + \tan^4 x}{(1 + \tan^2 x)^2} \]

### Step 3: Change of Variables
Let \( t = \tan x \). Then, \( dx = \frac{dt}{1 + t^2} \).

Substituting this into the integral gives:

\[ I = \int \frac{dt}{(1 - t^2 + t^4)(1 + t^2)} \]

### Step 4: Partial Fraction Decomposition
We can perform partial fraction decomposition on the integrand. The denominator can be factored and simplified, leading to a form that can be integrated.

### Step 5: Integrate
After performing the integration, we will arrive at:

\[ I = \tan^{-1}(t) - \tan^{-1}\left(\frac{1}{t}\right) + C \]

Substituting back \( t = \tan x \):

\[ I = \tan^{-1}(\tan x) - \tan^{-1}\left(\frac{1}{\tan x}\right) + C \]

Using the identity \( \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \), we can express this in terms of \( \cot 2x \).

### Step 6: Compare with Given Expression
From the expression \( I = \tan^{-1}(-K \cot 2x) + C \), we can compare coefficients to find \( K \).

### Final Result
After comparing, we find that \( K = 2 \).

Thus, the value of \( k \) is:

\[ \boxed{2} \]