Question

$\int(\sin^{-1}\sqrt{x}+\cos^{-1}\sqrt{x})dx=$ ________.

• A. $\frac{\pi}{2}+C$
• B. $\frac{\pi x}{4}+C$
• C. $\frac{\pi x}{3}+C$
• D. $\frac{\pi x}{2}+C$
• E. $\frac{-\pi x}{2}+C$

Hint:

Inverse trigonometric sums equal to $\pi/2$ are common simplifiers.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This problem involves integrating a sum of inverse trigonometric functions. Before attempting a complex integration, we should always check for identities that can simplify the integrand.
Step 2: Key Formula or Approach
We will use the fundamental inverse trigonometric identity:
\[ \sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2} \] This identity is valid for all \(u\) in the domain \([-1, 1]\).
Step 3: Detailed Explanation
1. Simplify the integrand.
The integrand is \(\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}\).
For this expression to be defined, the argument \(\sqrt{x}\) must be in the domain of both \(\sin^{-1}\) and \(\cos^{-1}\), which is \([-1, 1]\). Since \(\sqrt{x}\) cannot be negative, the domain is \(0 \le \sqrt{x} \le 1\), which implies \(0 \le x \le 1\).
Within this domain, we can apply the identity with \(u = \sqrt{x}\).
\[ \sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x} = \frac{\pi}{2} \] 2. Integrate the simplified expression.
The integral becomes:
\[ \int \frac{\pi}{2} \, dx \] Since \(\frac{\pi}{2}\) is a constant, we can pull it out of the integral.
\[ \frac{\pi}{2} \int 1 \, dx \] The integral of 1 with respect to \(x\) is \(x\).
\[ \frac{\pi}{2} x + C \] Step 4: Final Answer
The integral is \(\frac{\pi}{2}x + C\).