`int_0^(1)(sin^(- 1)x)/x dx=`

Options

A. `(pi)/(8) ln 2`
B. `(pi)/(4) ln 2`
C. `(pi)/(2 sqrt2) ln 2`
D. `(pi)/(2) ln 2`

Answer

To solve the integral \( I = \int_0^1 \frac{\sin^{-1} x}{x} \, dx \), we can use a substitution method and integration by parts. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = \sin^{-1} x \). Then, we have:
\[ x = \sin t \quad \text{and} \quad dx = \cos t \, dt \]
The limits change as follows:
- When \( x = 0 \), \( t = \sin^{-1}(0) = 0 \)
- When \( x = 1 \), \( t = \sin^{-1}(1) = \frac{\pi}{2} \)

### Step 2: Rewrite the Integral
Substituting these into the integral, we get:
\[ I = \int_0^{\frac{\pi}{2}} \frac{t}{\sin t} \cos t \, dt \]
This simplifies to:
\[ I = \int_0^{\frac{\pi}{2}} t \cdot \frac{\cos t}{\sin t} \, dt = \int_0^{\frac{\pi}{2}} t \cot t \, dt \]

### Step 3: Integration by Parts
Now, we will use integration by parts. Let:
- \( u = t \) (thus \( du = dt \))
- \( dv = \cot t \, dt \) (thus \( v = \ln(\sin t) \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[ I = \left[ t \ln(\sin t) \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \ln(\sin t) \, dt \]

### Step 4: Evaluate the Boundary Terms
Evaluating the boundary terms:
- At \( t = \frac{\pi}{2} \), \( \sin\left(\frac{\pi}{2}\right) = 1 \) so \( \ln(1) = 0 \).
- At \( t = 0 \), \( \sin(0) = 0 \) and \( t \ln(\sin t) \) approaches \( 0 \) as \( t \to 0 \) (using L'Hôpital's Rule).

Thus, the boundary terms contribute \( 0 - 0 = 0 \).

### Step 5: Final Integral
Now we have:
\[ I = - \int_0^{\frac{\pi}{2}} \ln(\sin t) \, dt \]
It is known that:
\[ \int_0^{\frac{\pi}{2}} \ln(\sin t) \, dt = -\frac{\pi}{2} \ln(2) \]
Therefore:
\[ I = -\left(-\frac{\pi}{2} \ln(2)\right) = \frac{\pi}{2} \ln(2) \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^1 \frac{\sin^{-1} x}{x} \, dx = \frac{\pi}{2} \ln(2) \]

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