Question

Evaluate \( \int \frac{\sin^{-1}(x)}{1 + x^2} \, dx \)

Hint:

For integrals involving inverse trigonometric functions, use substitutions and integration by parts to simplify the expressions.

Answer 1

Step 1: Using a substitution for inverse sine.
Let \( u = \sin^{-1}(x) \), so \( x = \sin(u) \). Differentiating both sides with respect to \( x \), we get: \[ \frac{dx}{du} = \cos(u) \quad \Rightarrow \quad dx = \cos(u) \, du \] Now, substitute into the integral. Since \( x = \sin(u) \), we know that \( 1 + x^2 = 1 + \sin^2(u) = \cos^2(u) \). Thus, the integral becomes: \[ I = \int \frac{u}{\cos^2(u)} \, du \] Step 2: Simplifying the expression.
We now simplify the integrand: \[ I = \int u \sec^2(u) \, du \] To integrate, use integration by parts, where: \[ \int u \sec^2(u) \, du = u \tan(u) - \int \tan(u) \, du \] Step 3: Solving the integral.
The integral of \( \tan(u) \) is \( -\ln|\cos(u)| \), so: \[ I = u \tan(u) + \ln|\cos(u)| + C \] Now, substitute \( u = \sin^{-1}(x) \) back into the result: \[ I = \sin^{-1}(x) \cdot \frac{x}{\sqrt{1 - x^2}} + \ln|\cos(\sin^{-1}(x))| + C \] This is the evaluated integral.