The value of `int(cos^4xdx)/(sin^3x(sin^5x+cos^5x)^(3//5))=-(1)/(2)(1+g(x))^(2//5)+C` where g (x) is :

Answer

To solve the integral

\[ \int \frac{\cos^4 x \, dx}{\sin^3 x \left(\sin^5 x + \cos^5 x\right)^{3/5}} = -\frac{1}{2} \left(1 + g(x)\right)^{2/5} + C \]

where \(g(x)\) is to be determined, we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral

\[ \int \frac{\cos^4 x \, dx}{\sin^3 x \left(\sin^5 x + \cos^5 x\right)^{3/5}}. \]

We can factor out \(\sin^5 x\) from the denominator:

\[ \sin^5 x + \cos^5 x = \sin^5 x \left(1 + \left(\frac{\cos x}{\sin x}\right)^5\right) = \sin^5 x \left(1 + \cot^5 x\right). \]

Thus, we rewrite the integral as:

\[ \int \frac{\cos^4 x \, dx}{\sin^3 x \cdot \sin^{15/5} x \cdot \left(1 + \cot^5 x\right)^{3/5}}. \]

### Step 2: Simplify the Integral
This simplifies to:

\[ \int \frac{\cos^4 x \, dx}{\sin^{18/5} x \cdot \left(1 + \cot^5 x\right)^{3/5}}. \]

Now, we can express \(\frac{1}{\sin^2 x}\) as \(\cot^2 x\):

\[ \int \cot^4 x \cdot \frac{\cos^2 x \, dx}{\sin^{18/5} x \cdot \left(1 + \cot^5 x\right)^{3/5}}. \]

### Step 3: Use Substitution
Let

\[ t = 1 + \cot^5 x. \]

Then, differentiate \(t\):

\[ dt = 5 \cot^4 x (-\csc^2 x) \, dx \implies dx = -\frac{dt}{5 \cot^4 x \csc^2 x}. \]

Substituting this back into the integral gives:

\[ -\frac{1}{5} \int \frac{1}{t^{3/5}} \, dt. \]

### Step 4: Integrate
The integral

\[ \int t^{-3/5} \, dt = \frac{t^{2/5}}{2/5} = \frac{5}{2} t^{2/5} + C. \]

Thus, we have:

\[ -\frac{1}{5} \cdot \frac{5}{2} t^{2/5} + C = -\frac{1}{2} t^{2/5} + C. \]

### Step 5: Substitute Back
Now substituting back for \(t\):

\[ -\frac{1}{2} (1 + \cot^5 x)^{2/5} + C. \]

### Step 6: Identify \(g(x)\)
From the original equation, we have:

\[ -\frac{1}{2} (1 + g(x))^{2/5} = -\frac{1}{2} (1 + \cot^5 x)^{2/5}. \]

Thus, we can equate:

\[ g(x) = \cot^5 x. \]

### Final Answer
The value of \(g(x)\) is:

\[ \boxed{\cot^5 x}. \]