Question

Evaluate: \[ \int \sin^{-1} x \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{1 - x^2}} \right) dx \]

Hint:

For integrals involving inverse trigonometric functions, consider using substitution and integration by parts to simplify the process.

Answer 1

Step 1: Simplify the integrand.
First, let's simplify the integrand \( \frac{x + \sqrt{1 - x^2}}{\sqrt{1 - x^2}} \). We can break this up as: \[ \frac{x}{\sqrt{1 - x^2}} + \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^2}} = \frac{x}{\sqrt{1 - x^2}} + 1 \] Thus, the integral becomes: \[ \int \sin^{-1}x \left( \frac{x}{\sqrt{1 - x^2}} + 1 \right) dx \]

Step 2: Split the integral.
We can now split the integral into two parts: \[ \int \sin^{-1}x \cdot \frac{x}{\sqrt{1 - x^2}} \, dx + \int \sin^{-1}x \, dx \]

Step 3: Solve the first integral.
We make the substitution \( u = \sin^{-1}x \), so that \( x = \sin u \) and \( dx = \cos u \, du \). The first integral becomes: \[ \int u \cdot \frac{\sin u}{\cos u} \cdot \cos u \, du = \int u \sin u \, du \] This can be solved by integration by parts. Let \( v = u \) and \( dw = \sin u \, du \), then: \[ \int u \sin u \, du = -u \cos u + \int \cos u \, du = -u \cos u + \sin u \]

Step 4: Solve the second integral.
The second integral is simply: \[ \int \sin^{-1}x \, dx = \int u \, du = \frac{u^2}{2} + C \]

Step 5: Combine the results.
Thus, the original integral becomes: \[ \boxed{-\sin^{-1}x \cos(\sin^{-1}x) + \sin(\sin^{-1}x) + \frac{(\sin^{-1}x)^2}{2} + C} \]