If f(x) = cos(tan^{-1}x),then the value of th | vedclass

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(D) Given $f(x) = \cos(	an^{-1}x)$.Using integration by parts for $I = \int_{0}^{1} x f''(x) dx$:$I = [x f'(x)]_{0}^{1} - \int_{0}^{1} f'(x) dx$$I =

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$1 - \frac{3}{2\sqrt{2}}$

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(D) Given $f(x) = \cos(	an^{-1}x)$.Using integration by parts for $I = \int_{0}^{1} x f''(x) dx$:$I = [x f'(x)]_{0}^{1} - \int_{0}^{1} f'(x) dx$$I = f'(1) - [f(1) - f(0)] = f'(1) - f(1) + f(0)$.Now,$f(x) = \cos(	an^{-1}x)$. At $x=1$,$	an^{-1}(1) = \frac{\pi}{4}$,so $f(1) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.At $x=0$,$	an^{-1}(0) = 0$,so $f(0) = \cos(0) = 1$.Derivative $f'(x) = -\sin(	an^{-1}x) \cdot \frac{1}{1+x^2}$.$f'(1) = -\sin(\frac{\pi}{4}) \cdot \frac{1}{1+1^2} = -\frac{1}{\sqrt{2}} \cdot \frac{1}{2} = -\frac{1}{2\sqrt{2}}$.Substituting these values:$I = -\frac{1}{2\sqrt{2}} - \frac{1}{\sqrt{2}} + 1 = 1 - \frac{1+2}{2\sqrt{2}} = 1 - \frac{3}{2\sqrt{2}}$.

If $f(x) = \cos(\tan^{-1}x)$,then the value of the integral $\int_{0}^{1} x f''(x) dx$ is

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