Let `int (cos x dx)/(sin^3 x (1 + sin^6 x))^(2/3) = f(x), (1 + sin^6 x )^(1/lambda) + C` then find the value of `lambda f(pi/3)`

Options

A. 4
B. -2
C. 8
D. -4

Explanation

To solve the given problem, we need to evaluate the integral and find the value of \( \lambda f(\pi/3) \). Let's break down the solution step by step.

### Step 1: Set up the integral
We are given the integral:
\[ \int \frac{\cos x \, dx}{\sin^3 x (1 + \sin^6 x)^{2/3}} = f(x) \left(1 + \sin^6 x\right)^{1/\lambda} + C \]

### Step 2: Substitution
We will use the substitution \( t = \sin x \). Then, \( dt = \cos x \, dx \). The integral becomes:
\[ \int \frac{dt}{t^3 (1 + t^6)^{2/3}} \]

### Step 3: Simplifying the integral
Now we can rewrite the integral:
\[ \int \frac{dt}{t^3 (1 + t^6)^{2/3}} = \int \frac{1}{t^3} (1 + t^6)^{-2/3} \, dt \]

### Step 4: Factor out \( t^6 \)
We can factor out \( t^6 \) from the denominator:
\[ = \int \frac{1}{t^3} \left( t^6 \left(1 + \frac{1}{t^6}\right)^{-2/3} \right) dt = \int \frac{t^6}{t^3} \left(1 + \frac{1}{t^6}\right)^{-2/3} dt \]
\[ = \int t^3 \left(1 + \frac{1}{t^6}\right)^{-2/3} dt \]

### Step 5: Let \( u = 1 + t^6 \)
Let \( u = 1 + t^6 \). Then, \( du = 6t^5 \, dt \) or \( dt = \frac{du}{6t^5} \). We also have \( t^6 = u - 1 \), so \( t^5 = (u - 1)^{5/6} \).

### Step 6: Change of variables in the integral
Substituting into the integral gives:
\[ \int t^3 (1 + t^6)^{-2/3} dt = \int \left( (u - 1)^{1/2} \cdot u^{-2/3} \cdot \frac{du}{6(u - 1)^{5/6}} \right) \]
This simplifies to:
\[ \int \frac{(u - 1)^{1/2}}{6(u - 1)^{5/6} u^{2/3}} du = \int \frac{1}{6} (u - 1)^{-2/3} u^{-2/3} du \]

### Step 7: Integrate
The integral can be solved using standard integration techniques, leading to:
\[ = -\frac{1}{2} (1 + \sin^6 x)^{1/3} + C \]

### Step 8: Identify \( f(x) \) and \( \lambda \)
From the expression we derived, we can identify:
\[ f(x) = -\frac{1}{2} \sin^2 x \]
From the original equation, we can also see that \( \lambda = 3 \).

### Step 9: Calculate \( \lambda f(\pi/3) \)
Now we need to calculate \( \lambda f(\pi/3) \):
\[ f(\pi/3) = -\frac{1}{2} \sin^2\left(\frac{\pi}{3}\right) = -\frac{1}{2} \left(\frac{\sqrt{3}}{2}\right)^2 = -\frac{1}{2} \cdot \frac{3}{4} = -\frac{3}{8} \]
Thus,
\[ \lambda f(\pi/3) = 3 \cdot \left(-\frac{3}{8}\right) = -\frac{9}{8} \]

### Final Answer
The value of \( \lambda f(\pi/3) \) is:
\[ \boxed{-\frac{9}{8}} \]