`lim_(trarr(pi/2)^(-))int_(0)^(t) tanthetasqrt(costheta)ln(costheta)d theta` is equal to :

Options

A. `-4`
B. `4`
C. `-2`
D. Does not exist

Answer

To solve the limit of the integral as \( t \) approaches \( \frac{\pi}{2} \) from the left, we start with the expression:

\[ \lim_{t \to \frac{\pi}{2}^-} \int_0^t \tan \theta \sqrt{\cos \theta} \ln(\cos \theta) \, d\theta \]

### Step 1: Set Up the Integral
Let:
\[ I = \int_0^t \tan \theta \sqrt{\cos \theta} \ln(\cos \theta) \, d\theta \]

### Step 2: Change of Variables
We will use the substitution \( u = \ln(\cos \theta) \). Then, we have:
\[ \cos \theta = e^u \quad \text{and} \quad d\theta = \frac{-\sin \theta}{\cos \theta} \, du = -\tan \theta \, du \]
This means:
\[ \tan \theta \, d\theta = -du \]

### Step 3: Express the Integral in Terms of \( u \)
Substituting into the integral, we get:
\[ I = \int_{u_0}^{u_t} -\sqrt{e^u} \cdot u \, du \]
where \( u_0 = \ln(1) = 0 \) (when \( \theta = 0 \)) and \( u_t = \ln(\cos t) \) (when \( \theta = t \)).

Thus, we can rewrite the integral as:
\[ I = -\int_{0}^{\ln(\cos t)} \sqrt{e^u} \cdot u \, du = -\int_{0}^{\ln(\cos t)} e^{u/2} u \, du \]

### Step 4: Integration by Parts
Using integration by parts, let:
- \( v = u \) and \( dv = du \)
- \( dw = e^{u/2} \, du \) and \( w = 2e^{u/2} \)

Then, we have:
\[ \int u e^{u/2} \, du = 2u e^{u/2} - \int 2 e^{u/2} \, du \]
The integral of \( 2 e^{u/2} \) is:
\[ 4 e^{u/2} \]
Thus,
\[ \int u e^{u/2} \, du = 2u e^{u/2} - 4 e^{u/2} + C \]

### Step 5: Evaluate the Integral
Now, substituting back, we evaluate:
\[ I = -\left[ 2u e^{u/2} - 4 e^{u/2} \right]_{0}^{\ln(\cos t)} \]
Calculating the limits:
- At \( u = \ln(\cos t) \):
\[ 2 \ln(\cos t) e^{\ln(\cos t)/2} - 4 e^{\ln(\cos t)/2} = 2 \ln(\cos t) \sqrt{\cos t} - 4 \sqrt{\cos t} \]
- At \( u = 0 \):
\[ 2(0)(1) - 4(1) = -4 \]

Thus, we have:
\[ I = -\left( 2 \ln(\cos t) \sqrt{\cos t} - 4 + 4 \right) = -2 \ln(\cos t) \sqrt{\cos t} \]

### Step 6: Take the Limit
Now, we need to evaluate:
\[ \lim_{t \to \frac{\pi}{2}^-} -2 \ln(\cos t) \sqrt{\cos t} \]
As \( t \to \frac{\pi}{2}^- \), \( \cos t \to 0^+ \), thus \( \ln(\cos t) \to -\infty \) and \( \sqrt{\cos t} \to 0 \). The product \( -2 \ln(\cos t) \sqrt{\cos t} \) approaches \( 0 \) because \( \sqrt{\cos t} \) approaches \( 0 \) faster than \( \ln(\cos t) \) approaches \( -\infty \).

### Final Answer
Thus, we conclude:
\[ \lim_{t \to \frac{\pi}{2}^-} I = 0 \]