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

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(C) Given $f^{\prime}(x)=	an^{-1}(\sec x+	an x)$.We simplify the expression inside the inverse tangent:$f^{\prime}(x)=	an^{-1}\left(\frac{1+\sin x}

Vedclass · Accepted Answer

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

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(C) Given $f^{\prime}(x)=	an^{-1}(\sec x+	an x)$.We simplify the expression inside the inverse tangent:$f^{\prime}(x)=	an^{-1}\left(\frac{1+\sin x}{\cos x}ight) = 	an^{-1}\left(\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}ight) = 	an^{-1}\left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}ight)$Dividing numerator and denominator by $\cos \frac{x}{2}$:$f^{\prime}(x)=	an^{-1}\left(\frac{1+	an \frac{x}{2}}{1-	an \frac{x}{2}}ight) = 	an^{-1}\left(	an(\frac{\pi}{4}+\frac{x}{2})ight)$Since $-\frac{\pi}{2} Integrating with respect to $x$:$f(x) = \int (\frac{\pi}{4} + \frac{x}{2}) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C$.Given $f(0)=0$,we find $C=0$.Thus,$f(x) = \frac{\pi x + x^2}{4}$.For $x=1$,$f(1) = \frac{\pi(1) + (1)^2}{4} = \frac{\pi+1}{4}$.

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

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