Q112 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

$\int e^{-2 x}\left(\tan 2 x-2 \sec ^2 2 x \tan 2 x\right) d x=$

  • A
    $e^{-2 x} \tan 2 x+c$
  • Correct answer
    B
    $-\frac{e^{-2 x}}{2}\left[\sec ^2 2 x+\tan 2 x\right]+c$
  • C
    $-\frac{e^{-2 x}}{2}\left[\tan 2 x-\sec ^2 2 x\right]+c$
  • D
    $e^{-2 x} \sec ^2 2 x+c$

Solution

Let $-2 x=t \Rightarrow-2 d x=d t$ $\begin{aligned} & \Rightarrow d x=-\frac{1}{2} d t=-\frac{1}{2} \int e^t\left(-\tan t+2 \sec ^2 t \tan t\right) d t \\ & =-\frac{1}{2} \int e^t\left[\sec ^2 t-\tan t+2 \sec ^2 t \tan t-\sec ^2 t\right] d t \\ & \because \frac{d\left(\sec ^2 t-\tan t\right)}{d t}=2 \sec ^2 t \tan t-\sec ^2 t \\ & \therefore I=\frac{-1}{2} e^t\left[\sec ^2 t-\tan t\right]=\frac{-1}{2} e^{-2 x}\left[\sec ^2 2 x+\tan 2 x\right]+C\end{aligned}$

Hence, the correct answer is option: 'B'.

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