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If $\mathrm{f}^{\prime}(\mathrm{x})=\tan ^{-1}(\sec \mathrm{x} \tan \mathrm{x}), \frac{-\pi}{2}<\mathrm{x}<\frac{\pi}{2}$ and $\mathrm{f}(0)=0$, then $\mathrm{f}(1)=$
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Verified Answer
The correct answer is:
$\frac{\pi+1}{4}$
$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\tan ^{-1}(\sec \mathrm{x}+\tan \mathrm{x})=\tan ^{-1}\left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}\right) \\ & =\tan ^{-1}\left(\frac{\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2}}{\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2}}\right)=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\right) \\ & =\frac{\pi}{4}+\frac{\mathrm{x}}{2}\end{aligned}$
Now $f(x)=\int f^{\prime}(x) d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x=\frac{\pi}{4} x+\frac{x^2}{4}+C$
$$
\because \mathrm{f}(0)=0 \Rightarrow \mathrm{c}=0
$$
Hence, $f(x)=\frac{x^2+\pi x}{4}$
$$
\Rightarrow \mathrm{f}(1)=\frac{1^2+\pi \times 1}{4}=\frac{\pi+1}{4}
$$
Now $f(x)=\int f^{\prime}(x) d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x=\frac{\pi}{4} x+\frac{x^2}{4}+C$
$$
\because \mathrm{f}(0)=0 \Rightarrow \mathrm{c}=0
$$
Hence, $f(x)=\frac{x^2+\pi x}{4}$
$$
\Rightarrow \mathrm{f}(1)=\frac{1^2+\pi \times 1}{4}=\frac{\pi+1}{4}
$$
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