If f(x) = sin(tan^{-1} x),then int_0^1 x f''( | vedclass

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(B) Given $f(x) = \sin(	an^{-1} x)$.Using the identity $\sin(	an^{-1} x) = \frac{x}{\sqrt{1+x^2}}$,we have $f(x) = \frac{x}{\sqrt{1+x^2}}$.We need t

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$-\frac{1}{2\sqrt{2}}$

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(B) Given $f(x) = \sin(	an^{-1} x)$.Using the identity $\sin(	an^{-1} x) = \frac{x}{\sqrt{1+x^2}}$,we have $f(x) = \frac{x}{\sqrt{1+x^2}}$.We need to evaluate $I = \int_0^1 x f''(x) dx$.Using integration by parts,let $u = x$ and $dv = f''(x) dx$. Then $du = dx$ and $v = f'(x)$.$I = [x f'(x)]_0^1 - \int_0^1 f'(x) dx$.$I = (1 \cdot f'(1) - 0 \cdot f'(0)) - [f(1) - f(0)]$.First,calculate $f'(x) = \frac{d}{dx} \left( \frac{x}{\sqrt{1+x^2}} ight) = \frac{\sqrt{1+x^2} - x \cdot \frac{x}{\sqrt{1+x^2}}}{1+x^2} = \frac{1+x^2-x^2}{(1+x^2)^{3/2}} = \frac{1}{(1+x^2)^{3/2}}$.Now,$f'(1) = \frac{1}{(1+1^2)^{3/2}} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}}$.Also,$f(1) = \sin(	an^{-1} 1) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $f(0) = \sin(	an^{-1} 0) = 0$.Substituting these values into the expression for $I$:$I = \frac{1}{2\sqrt{2}} - (\frac{1}{\sqrt{2}} - 0) = \frac{1}{2\sqrt{2}} - \frac{2}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}}$.

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

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