Let `intsin(2x)ln(cosx)dx=f(x)cos^(2)x+C`, (where, C is the constant of integration) and `f(0)=(1)/(2)`, If `f((pi)/(3))` is equal to `(1)/(a)+lnb,` then the value of `a+b` is

Answer

To solve the given problem step by step, we will follow the integration and substitution method as described in the video transcript.

### Step 1: Set up the integral
We start with the integral:
\[ I = \int \sin(2x) \ln(\cos x) \, dx \]

### Step 2: Use the double angle formula
Using the double angle formula for sine, we have:
\[ \sin(2x) = 2 \sin x \cos x \]
Thus, we can rewrite the integral as:
\[ I = \int 2 \sin x \cos x \ln(\cos x) \, dx \]

### Step 3: Substitute for \(\cos x\)
Let \(t = \cos x\), then \(dt = -\sin x \, dx\) or \(-dt = \sin x \, dx\). Therefore, we can rewrite the integral as:
\[ I = -2 \int t \ln(t) \, dt \]

### Step 4: Integration by parts
Using integration by parts where \(u = \ln(t)\) and \(dv = t \, dt\):
- Then \(du = \frac{1}{t} dt\) and \(v = \frac{t^2}{2}\).
Using the integration by parts formula \(\int u \, dv = uv - \int v \, du\):
\[ I = -2 \left( \frac{t^2}{2} \ln(t) - \int \frac{t^2}{2} \cdot \frac{1}{t} dt \right) \]
\[ = -t^2 \ln(t) + \int t \, dt \]
\[ = -t^2 \ln(t) + \frac{t^2}{2} + C \]

### Step 5: Substitute back for \(t\)
Substituting back \(t = \cos x\):
\[ I = -\cos^2(x) \ln(\cos x) + \frac{\cos^2(x)}{2} + C \]

### Step 6: Relate to \(f(x)\)
We know from the problem statement:
\[ I = f(x) \cos^2(x) + C \]
Thus, we can equate:
\[ f(x) \cos^2(x) = -\cos^2(x) \ln(\cos x) + \frac{\cos^2(x)}{2} \]
Dividing both sides by \(\cos^2(x)\) (assuming \(\cos x \neq 0\)):
\[ f(x) = -\ln(\cos x) + \frac{1}{2} \]

### Step 7: Find \(f(0)\)
To find \(f(0)\):
\[ f(0) = -\ln(\cos(0)) + \frac{1}{2} = -\ln(1) + \frac{1}{2} = \frac{1}{2} \]
This confirms \(f(0) = \frac{1}{2}\).

### Step 8: Find \(f\left(\frac{\pi}{3}\right)\)
Now, we need to calculate \(f\left(\frac{\pi}{3}\right)\):
\[ f\left(\frac{\pi}{3}\right) = -\ln\left(\cos\left(\frac{\pi}{3}\right)\right) + \frac{1}{2} \]
Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\):
\[ f\left(\frac{\pi}{3}\right) = -\ln\left(\frac{1}{2}\right) + \frac{1}{2} = \ln(2) + \frac{1}{2} \]

### Step 9: Set up the equation
According to the problem:
\[ f\left(\frac{\pi}{3}\right) = \frac{1}{a} + \ln(b) \]
Comparing:
\[ \frac{1}{2} + \ln(2) = \frac{1}{a} + \ln(b) \]

### Step 10: Identify \(a\) and \(b\)
From the comparison:
1. \(\frac{1}{a} = \frac{1}{2} \implies a = 2\)
2. \(\ln(b) = \ln(2) \implies b = 2\)

### Step 11: Calculate \(a + b\)
Finally, we find:
\[ a + b = 2 + 2 = 4 \]

Thus, the final answer is:
\[ \boxed{4} \]