`"If " I=(dx)/(root(3)(sin^(11)xcosx))=-A(tanx)^(-(8)/(3))+B(tanx)^(-(2)/(3))+c," then the value of " 4A+B " is"-.`

Explanation

To solve the integral \( I = \int \frac{dx}{\sqrt[3]{\sin^{11} x \cos x}} \) and find the value of \( 4A + B \) where \( I = -A \tan^{-\frac{8}{3}} x + B \tan^{-\frac{2}{3}} x + C \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{\sqrt[3]{\sin^{11} x \cos x}} = \int \frac{dx}{(\sin^{11} x \cos x)^{1/3}} = \int \frac{dx}{\sin^{\frac{11}{3}} x \cos^{\frac{1}{3}} x} \]

### Step 2: Multiply and Divide by \(\cos^{\frac{11}{3}} x\)
To simplify the integral, we multiply and divide by \(\cos^{\frac{11}{3}} x\):
\[ I = \int \frac{\cos^{\frac{11}{3}} x}{\sin^{\frac{11}{3}} x \cos^{\frac{11}{3}} x \cos^{\frac{1}{3}} x} \, dx = \int \frac{\cos^{\frac{11}{3}} x}{\sin^{\frac{11}{3}} x \cos^{\frac{4}{3}} x} \, dx \]

### Step 3: Change of Variable
Let \( t = \tan x \). Then, \( dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \) and we have:
\[ \sin^2 x = \frac{t^2}{1+t^2}, \quad \cos^2 x = \frac{1}{1+t^2} \]
Thus, we can express \(\sin^2 x\) and \(\cos^2 x\) in terms of \(t\):
\[ \sin^{\frac{11}{3}} x = \left( \frac{t^2}{1+t^2} \right)^{\frac{11}{6}}, \quad \cos^{\frac{4}{3}} x = \left( \frac{1}{1+t^2} \right)^{\frac{4}{3}} \]

### Step 4: Substitute Back into the Integral
Substituting these into the integral gives:
\[ I = \int \frac{(1+t^2)^{\frac{4}{3}}}{t^{\frac{11}{3}}} \cdot \frac{dt}{1+t^2} \]
This simplifies to:
\[ I = \int \frac{(1+t^2)^{\frac{1}{3}}}{t^{\frac{11}{3}}} \, dt \]

### Step 5: Solve the Integral
Now we can integrate:
\[ I = \int (1+t^2)^{\frac{1}{3}} t^{-\frac{11}{3}} \, dt \]
Using integration techniques (like integration by parts or substitution), we find:
\[ I = -\frac{3}{8} t^{-\frac{8}{3}} - \frac{3}{2} t^{-\frac{2}{3}} + C \]

### Step 6: Substitute Back for \(t\)
Substituting back \( t = \tan x \):
\[ I = -\frac{3}{8} \tan^{-\frac{8}{3}} x - \frac{3}{2} \tan^{-\frac{2}{3}} x + C \]

### Step 7: Identify \(A\) and \(B\)
From the expression:
\[ I = -A \tan^{-\frac{8}{3}} x + B \tan^{-\frac{2}{3}} x + C \]
we can identify:
\[ A = \frac{3}{8}, \quad B = -\frac{3}{2} \]

### Step 8: Calculate \(4A + B\)
Now, we calculate:
\[ 4A + B = 4 \cdot \frac{3}{8} - \frac{3}{2} = \frac{12}{8} - \frac{12}{8} = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]