Find the general solution of the following :
`(dy)/(dx)= (1)/(sin^(4)x+cos^(4)x)`

Answer

To find the general solution of the differential equation

\[ \frac{dy}{dx} = \frac{1}{\sin^4 x + \cos^4 x}, \]

we will proceed with the following steps:

### Step 1: Rewrite the equation
We start by rewriting the differential equation:

\[ dy = \frac{1}{\sin^4 x + \cos^4 x} \, dx. \]

### Step 2: Integrate both sides
Next, we integrate both sides:

\[ \int dy = \int \frac{1}{\sin^4 x + \cos^4 x} \, dx. \]

The left side simplifies to:

\[ y = \int \frac{1}{\sin^4 x + \cos^4 x} \, dx + C, \]

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

### Step 3: Simplify the integrand
To simplify \(\sin^4 x + \cos^4 x\), we can use the identity:

\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x. \]

Using \(\sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x)\), we have:

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

### Step 4: Substitute and integrate
Now we substitute this back into the integral:

\[ y = \int \frac{1}{1 - \frac{1}{2} \sin^2(2x)} \, dx + C. \]

This can be rewritten as:

\[ y = \int \frac{2}{2 - \sin^2(2x)} \, dx + C. \]

### Step 5: Use a trigonometric identity
To solve this integral, we can use the substitution \(u = 2x\), which gives \(du = 2dx\) or \(dx = \frac{du}{2}\):

\[ y = \int \frac{2}{2 - \sin^2(u)} \cdot \frac{du}{2} + C = \int \frac{1}{2 - \sin^2(u)} \, du + C. \]

### Step 6: Solve the integral
The integral \(\int \frac{1}{2 - \sin^2(u)} \, du\) can be solved using partial fractions or trigonometric identities. The result will involve the tangent inverse function.

### Step 7: Back substitute
After finding the integral, we will back substitute \(u = 2x\) to express the solution in terms of \(x\).

### Final Answer
The final answer will be in the form:

\[ y = \text{expression involving } \tan^{-1} \text{ and constants} + C. \]