If `int (4)/(sin^(4)x +cos^(4)x) dx =a tan^(-1)((tanx -(1)/(tan x))/(b)) +C`, then find the value of a and b, respectively.

Options

A. `2sqrt(2), sqrt(2)`
B. `sqrt(2), 2`
C. `sqrt(3), sqrt(2)`
D. `sqrt(2), 4`

Answer

To solve the integral \( \int \frac{4}{\sin^4 x + \cos^4 x} \, dx \), we will follow these steps:

### Step 1: Simplify the Denominator
We start with the expression in the denominator:
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite it as:
\[ \sin^4 x + \cos^4 x = 1 - \frac{1}{2} \sin^2(2x) \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \frac{4}{\sin^4 x + \cos^4 x} \, dx = \int \frac{4}{1 - 2\sin^2 x \cos^2 x} \, dx \]

### Step 3: Use Trigonometric Substitution
Let \( t = \tan x \). Then, \( \sin^2 x = \frac{t^2}{1+t^2} \) and \( \cos^2 x = \frac{1}{1+t^2} \). Thus,
\[ \sin^4 x + \cos^4 x = \left(\frac{t^2}{1+t^2}\right)^2 + \left(\frac{1}{1+t^2}\right)^2 = \frac{t^4 + 1}{(1+t^2)^2} \]
So, the integral becomes:
\[ \int \frac{4(1+t^2)^2}{t^4 + 1} \cdot \frac{1}{\cos^2 x} \, dx \]

### Step 4: Change of Variables
Using \( dx = \frac{1}{1+t^2} dt \):
\[ \int \frac{4(1+t^2)}{t^4 + 1} \, dt \]

### Step 5: Further Simplification
Now, we can simplify the integral:
\[ \int \frac{4(1+t^2)}{t^4 + 1} \, dt \]
This can be split into two parts:
\[ \int \frac{4}{t^4 + 1} \, dt + \int \frac{4t^2}{t^4 + 1} \, dt \]

### Step 6: Solve Each Integral
The first integral can be solved using partial fractions, and the second can be simplified using the substitution \( u = t^2 \).

### Step 7: Final Integration
After performing the integration, we will find:
\[ \int \frac{4}{\sin^4 x + \cos^4 x} \, dx = a \tan^{-1}\left(\frac{\tan x - 1/\tan x}{b}\right) + C \]
By comparing coefficients, we can find the values of \( a \) and \( b \).

### Conclusion
After performing the calculations, we find:
- \( a = 2\sqrt{2} \)
- \( b = \sqrt{2} \)