Q115 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics
$\int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec (2023 x)}{1+(2023)^{(2023 x)}} d x=$
- A$\frac{1}{2023 \sqrt{2}}+c$
- Correct answerB$\frac{\log (\sqrt{2}+1)}{2023}+c$
- C$\frac{\log 2}{4046}+c$
- D$\frac{\sqrt{2}}{2023}+c$
Solution
$I=\int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec 2023 x}{1+(2023)^{(2023 x)}} d x$
Using $f(x) \rightarrow f(a+b-x)$ ...(1)
$$
\Rightarrow I=\int_{\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec 2023 x}{1+(2023)^{-2023 x}} d x
$$
$\Rightarrow I=\int_{\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{(2023)^{2023 x} \cdot \sec 2023 x}{(2023)^{2023 x}+1} d x$ ...(2)
$\begin{aligned} & \text { Equations }(1)+(2), \\ & \Rightarrow 2 I=\int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \sec (2023 x) d x \\ & \Rightarrow 2 I=2 \int_0^{\frac{\pi}{8092}} \sec (2023 x) d x \\ & \Rightarrow I=\int_0^{\frac{\pi}{8092}} \sec (2023 x) d x, \quad(\because \sec \theta \text { is even function }) \\ & =\left[\log \frac{\sec (2023 x)+\tan (2023 x)}{2023}\right]_0^{\pi / 8092} \\ & =\frac{1}{2023}\left\{\log \left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)-\log \{\sec 0+\tan 0\}+C\right. \\ & =\frac{\log (\sqrt{2}+1)}{2023}+C .\end{aligned}$
Hence, the correct answer is option: 'B'.