Q116 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
$\int_{\pi / 5}^{3 \pi / 10} \frac{d x}{\sec ^2 x+\left(\tan ^{2022} x-1\right)\left(\sec ^2 x-1\right)}=$
- Correct answerA$\frac{\pi}{20}$
- B$\frac{2 \pi}{5}$
- C$\frac{3 \pi}{20}$
- D$\frac{3 \pi}{5}$
Solution
$I=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{d x}{1+\tan ^2 x+\left(\tan ^{2022} x-1\right) \tan ^2 x}$
$\Rightarrow I=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{d x}{1+\tan ^{2024} x}$ ...(i)
$\Rightarrow y=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{d x}{1+\tan ^{2024}\left(\frac{3 \pi}{10}+\frac{\pi}{5}-x\right)}$
$\Rightarrow I=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{d x}{1+\cot ^{2024} x} \Rightarrow I=\int \frac{\tan ^{2024} x}{1+\tan ^{2024} x} d x$ ...(ii)
Adding (i) and (ii)
$2 I=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\left(1+\tan ^{2024} x\right)}{\left(1+\tan ^{2024} x\right)} d x=\frac{3 \pi}{10}-\frac{\pi}{5} \quad \therefore I=\frac{\pi}{20} .$
Hence, the correct answer is option: 'A'.