If $\mathrm{f}^{\prime}(\mathrm{x})= an ^{-1}(\sec \mathrm{x} + an \mathrm{x}), \frac{-\pi}{2}

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=	an ^{-1}(\sec \mathrm{x}+	an \mathrm{x})=	an ^{-1}\left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}ight) \ & =	an ^{-1}\left(\frac{\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2}}{\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2}}ight)=	an ^{-1}\left(	an \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}ight)ight) \ & =\frac{\pi}{4}+\frac{\mathrm{x}}{2}\end{aligned}$ Now $f(x)=\int f^{\prime}(x) d x=\int\left(\frac{\pi}{4}+\frac{x}{2}ight) d x=\frac{\pi}{4} x+\frac{x^2}{4}+C$ $\because \mathrm{f}(0)=0 \Rightarrow \mathrm{c}=0$ Hence, $f(x)=\frac{x^2+\pi x}{4}$ $\Rightarrow \mathrm{f}(1)=\frac{1^2+\pi 	imes 1}{4}=\frac{\pi+1}{4}$

If $\mathrm{f}^{\prime}(\mathrm{x})=\tan ^{-1}(\sec \mathrm{x} + \tan \mathrm{x}),…

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