Step 1: Understanding the Concept:
To integrate inverse trigonometric functions alone, we treat the function as being multiplied by 1 and use integration by parts (ILATE rule).
: Key Formula or Approach:
Integration by parts: \( \int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx \).
Let \( u = \sin^{-1} x \) and \( v = 1 \).
Step 2: Detailed Explanation:
First, find the indefinite integral:
\[ \int \sin^{-1} x \cdot 1 \, dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^{2}}} dx \]
For the second term, use substitution \( 1-x^{2} = t \), then \( -2x dx = dt \):
\[ \int \frac{x}{\sqrt{1 - x^{2}}} dx = -\sqrt{1 - x^{2}} \]
So, \( \int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^{2}} \).
Now apply the limits from 0 to 1:
\[ I = \left[ x \sin^{-1} x + \sqrt{1 - x^{2}} \right]_{0}^{1} \]
\[ I = (1 \cdot \sin^{-1}(1) + \sqrt{1 - 1^{2}}) - (0 \cdot \sin^{-1}(0) + \sqrt{1 - 0^{2}}) \]
\[ I = (\pi/2 + 0) - (0 + 1) = \pi/2 - 1 \].
Step 3: Final Answer:
The value of the integral is \( \pi/2 - 1 \).