Question

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

• A. \(1-\dfrac{3}{2\sqrt2}\)
• B. \(-\dfrac{1}{2\sqrt2}\)
• C. \(\dfrac{1}{\sqrt2}\)
• D. \(-\sqrt2\)

Hint:

For integrals involving \(xf''(x)\), integration by parts is useful: \[ \int xf''(x)\,dx=xf'(x)-\int f'(x)\,dx. \]

Answer 1

The Correct Option is B

Step 1: Simplify \(f(x)\).
Let \[ \theta=\tan^{-1}x. \] Then, \[ \tan\theta=x. \] So, \[ \sin\theta=\frac{x}{\sqrt{1+x^2}}. \] Hence, \[ f(x)=\frac{x}{\sqrt{1+x^2}}. \]

Step 2: Differentiate \(f(x)\).
\[ f(x)=x(1+x^2)^{-1/2}. \] \[ f'(x)=(1+x^2)^{-1/2}-x^2(1+x^2)^{-3/2}. \] \[ f'(x)=\frac{1}{(1+x^2)^{3/2}}. \]

Step 3: Use integration by parts.
We need \[ \int_0^1 xf''(x)\,dx. \] Using integration by parts, \[ \int_0^1 xf''(x)\,dx = \left[xf'(x)\right]_0^1-\int_0^1 f'(x)\,dx. \] Now, \[ \left[xf'(x)\right]_0^1 = \frac{1}{(1+1^2)^{3/2}} = \frac{1}{2\sqrt2}. \] Also, \[ \int_0^1 f'(x)\,dx=f(1)-f(0). \] \[ f(1)=\frac{1}{\sqrt2},\qquad f(0)=0. \] So, \[ \int_0^1 f'(x)\,dx=\frac{1}{\sqrt2}. \]

Step 4: Calculate the value.
\[ \int_0^1 xf''(x)\,dx = \frac{1}{2\sqrt2}-\frac{1}{\sqrt2} \] \[ = -\frac{1}{2\sqrt2}. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{-\frac{1}{2\sqrt2}} \]