If `f(x)=sinx, g(x)=cosx and h(x)=cos(cosx),` then the integral `I=int f(g(x)).f(x).h(x)dx` simplifies to `-lambda sin^(2)(cosx)+C` (where, C is the constant of integration). The value of `lambda` is equal to

Answer

To solve the integral \( I = \int f(g(x)) f(x) h(x) \, dx \) where \( f(x) = \sin x \), \( g(x) = \cos x \), and \( h(x) = \cos(\cos x) \), we will follow these steps:

### Step 1: Substitute the functions
First, we substitute the definitions of \( f \), \( g \), and \( h \) into the integral:

\[ I = \int f(g(x)) f(x) h(x) \, dx = \int f(\cos x) \sin x \cos(\cos x) \, dx \]

### Step 2: Evaluate \( f(g(x)) \)
Since \( g(x) = \cos x \), we have:

\[ f(g(x)) = f(\cos x) = \sin(\cos x) \]

Thus, the integral becomes:

\[ I = \int \sin(\cos x) \sin x \cos(\cos x) \, dx \]

### Step 3: Use substitution
Let \( t = \sin(\cos x) \). Then we differentiate \( t \):

\[ dt = \cos(\cos x)(-\sin x) \, dx \implies -\sin x \cos(\cos x) \, dx = dt \]

This means \( dx = \frac{dt}{-\sin x \cos(\cos x)} \).

### Step 4: Substitute in the integral
Now we can rewrite the integral in terms of \( t \):

\[ I = \int t \left(-dt\right) = -\int t \, dt \]

### Step 5: Integrate
The integral of \( t \) is:

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

### Step 6: Substitute back for \( t \)
Substituting back \( t = \sin(\cos x) \):

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

### Step 7: Compare with the given form
The problem states that \( I \) simplifies to \( -\lambda \sin^2(\cos x) + C \). From our result:

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

We can see that:

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

### Step 8: Solve for \( \lambda \)
Thus, we find:

\[ \lambda = \frac{1}{2} \]

### Final Answer
The value of \( \lambda \) is \( \frac{1}{2} \).

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