Question

If \( f(x) = \sin(\tan^{-1} x) \), then \( \int_0^1 xf''(x)dx = \)

• A. \( 1 - \frac{3}{2\sqrt{2}} \)
• B. \( -\frac{1}{2\sqrt{2}} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( -\sqrt{2} \)

Hint:

Simplify \(f(x) = \sin(\tan^{-1} x)\) using a right-angled triangle. If \(\tan\theta = x/1\), then hypotenuse is \(\sqrt{x^2+1}\), so \(\sin\theta = x/\sqrt{x^2+1}\).
Use integration by parts: \( \int u dv = uv - \int v du \). Choose \(u=x\) and \(dv=f''(x)dx\).
Carefully calculate derivatives and evaluate at the limits.

Answer 1

The Correct Option is B

Let \( I = \int_0^1 xf''(x)dx \). We use integration by parts: \( \int u dv = uv - \int v du \). Let \(u = x \Rightarrow du = dx\). Let \(dv = f''(x)dx \Rightarrow v = f'(x)\). So, \( I = [x f'(x)]_0^1 - \int_0^1 f'(x)dx \). \( I = (1 \cdot f'(1) - 0 \cdot f'(0)) - [f(x)]_0^1 = f'(1) - (f(1) - f(0)) \). Given \( f(x) = \sin(\tan^{-1} x) \). Let \(\theta = \tan^{-1} x\), so \(\tan\theta = x\). Consider a right triangle with opposite side \(x\), adjacent side 1. Hypotenuse = \(\sqrt{x^2+1}\). Then \(\sin\theta = \frac{x}{\sqrt{x^2+1}}\). So, \(f(x) = \frac{x}{\sqrt{x^2+1}}\). Calculate \(f(0)\) and \(f(1)\): \(f(0) = \frac{0}{\sqrt{0^2+1}} = 0\). \(f(1) = \frac{1}{\sqrt{1^2+1}} = \frac{1}{\sqrt{2}}\). Calculate \(f'(x)\). \(f(x) = x(x^2+1)^{-1/2}\). Using quotient rule or product rule: \(f'(x) = \frac{1 \cdot \sqrt{x^2+1} - x \cdot \frac{1}{2\sqrt{x^2+1}}(2x)}{(x^2+1)} = \frac{\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}}{x^2+1}\) \(f'(x) = \frac{\frac{(x^2+1) - x^2}{\sqrt{x^2+1}}}{x^2+1} = \frac{1}{(x^2+1)\sqrt{x^2+1}} = (x^2+1)^{-3/2}\). Calculate \(f'(1)\): \(f'(1) = (1^2+1)^{-3/2} = (2)^{-3/2} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}}\). Substitute these values into the expression for I: \( I = f'(1) - (f(1) - f(0)) = \frac{1}{2\sqrt{2}} - \left(\frac{1}{\sqrt{2}} - 0\right) \) \( I = \frac{1}{2\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}} - \frac{2}{2\sqrt{2}} = \frac{1-2}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}} \). This matches option (b). \[ \boxed{-\frac{1}{2\sqrt{2}}} \]