`int(tanx)/(sqrt(sin^(4)x+cos^(4)x))dx` is equal to

Options

A. `log_(e)(tan^(2)x+sqrt(1+tan^(4)x))+C`
B. `(1)/(2)log_(e)(tan^(2)x+sqrt(1+tan^(4)x))+C`
C. `(1)/(4)log(tan^(2)x+sqrt(1+tan^(4)x))+C`
D. none of these

Explanation

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

### Step 1: Simplify the Denominator
We start by simplifying the expression under the square root in the denominator:
\[ \sin^4 x + \cos^4 x \]
Using the identity \( a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 \), we can rewrite it as:
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Thus, we have:
\[ \sqrt{\sin^4 x + \cos^4 x} = \sqrt{1 - 2\sin^2 x \cos^2 x} \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \frac{\tan x}{\sqrt{1 - 2\sin^2 x \cos^2 x}} \, dx \]
Since \( \tan x = \frac{\sin x}{\cos x} \), we can express the integral as:
\[ \int \frac{\sin x}{\cos x \sqrt{1 - 2\sin^2 x \cos^2 x}} \, dx \]

### Step 3: Use a Substitution
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \). Also, \( \sin x = \frac{t}{\sqrt{1+t^2}} \) and \( \cos x = \frac{1}{\sqrt{1+t^2}} \).

Substituting these into the integral gives:
\[ \int \frac{\frac{t}{\sqrt{1+t^2}}}{\frac{1}{\sqrt{1+t^2}} \sqrt{1 - 2\left(\frac{t^2}{1+t^2}\right)\left(\frac{1}{1+t^2}\right)}} \cdot \frac{dt}{1+t^2} \]
This simplifies to:
\[ \int \frac{t}{\sqrt{1 - \frac{2t^2}{(1+t^2)^2}}} \cdot \frac{dt}{1+t^2} \]

### Step 4: Further Simplification
Now, we need to simplify the expression under the square root:
\[ 1 - \frac{2t^2}{(1+t^2)^2} = \frac{(1+t^2)^2 - 2t^2}{(1+t^2)^2} = \frac{1 + 2t^2 + t^4 - 2t^2}{(1+t^2)^2} = \frac{1 + t^4}{(1+t^2)^2} \]
Thus, we have:
\[ \sqrt{1 - \frac{2t^2}{(1+t^2)^2}} = \frac{\sqrt{1 + t^4}}{1+t^2} \]

### Step 5: Substitute Back into the Integral
The integral now becomes:
\[ \int \frac{t(1+t^2)}{\sqrt{1+t^4}} \cdot \frac{dt}{1+t^2} = \int \frac{t}{\sqrt{1+t^4}} \, dt \]

### Step 6: Solve the Integral
This integral can be solved using the substitution \( u = 1 + t^4 \), leading to:
\[ \int \frac{t}{\sqrt{u}} \cdot \frac{du}{4t^3} = \frac{1}{4} \int u^{-1/2} \, du = \frac{1}{2} \sqrt{u} + C = \frac{1}{2} \sqrt{1 + t^4} + C \]

### Step 7: Substitute Back for \( t \)
Finally, substituting back \( t = \tan x \):
\[ \frac{1}{2} \sqrt{1 + \tan^4 x} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\tan x}{\sqrt{\sin^4 x + \cos^4 x}} \, dx = \frac{1}{2} \sqrt{1 + \tan^4 x} + C \]