Question

Integrate \[ \int \frac{\sin(\tan^{-1} x)}{1 + x^2} \, dx. \] 

Hint:

For integrals involving inverse trigonometric functions, use appropriate substitutions and standard identities to simplify the integrals.

Answer 1

Step 1: Use the substitution \( \theta = \tan^{-1}(x) \), so \( \tan(\theta) = x \) and \( \frac{d\theta}{dx} = \frac{1}{1 + x^2} \). Step 2: From the identity \( \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} \), we have: \[ \int \frac{\sin(\tan^{-1} x)}{1 + x^2} \, dx = \int \frac{x}{\sqrt{1 + x^2}} \cdot \frac{1}{1 + x^2} \, dx. \] Step 3: Simplify the integrand: \[ = \int \frac{x}{(1 + x^2)^{3/2}} \, dx. \] Step 4: Use the substitution \( u = 1 + x^2 \), so \( du = 2x \, dx \). This gives: \[ = \frac{1}{2} \int u^{-3/2} \, du. \] Step 5: Integrate: \[ = -\frac{1}{\sqrt{u}} + C = -\frac{1}{\sqrt{1 + x^2}} + C. \] Thus, the result is: \[ -\frac{1}{\sqrt{1 + x^2}} + C. \]