If f^{prime}(x) = tan^{-1}(sec x + tan x),fra | vedclass

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(C) Given $f^{\prime}(x) = 	an^{-1}(\sec x + 	an x) = 	an^{-1}\left(\frac{1+\sin x}{\cos x}ight)$.Using trigonometric identities,$\frac{1+\sin x}

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$\frac{\pi+1}{4}$

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(C) Given $f^{\prime}(x) = 	an^{-1}(\sec x + 	an x) = 	an^{-1}\left(\frac{1+\sin x}{\cos x}ight)$.Using trigonometric identities,$\frac{1+\sin x}{\cos x} = \frac{(\cos(x/2) + \sin(x/2))^2}{\cos^2(x/2) - \sin^2(x/2)} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} = 	an(\frac{\pi}{4} + \frac{x}{2})$.Thus,$f^{\prime}(x) = 	an^{-1}(	an(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$.Now,$f(x) = \int f^{\prime}(x) dx = \int (\frac{\pi}{4} + \frac{x}{2}) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C$.Since $f(0) = 0$,we have $0 = 0 + 0 + C$,so $C = 0$.Therefore,$f(x) = \frac{x^2 + \pi x}{4}$.Substituting $x = 1$,we get $f(1) = \frac{1^2 + \pi(1)}{4} = \frac{\pi+1}{4}$.

If $f^{\prime}(x) = \tan^{-1}(\sec x + \tan x)$,$\frac{-\pi}{2} < x < \frac{\pi}{2}$ and $f(0) = 0$,then $f(1) =$

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