Q88 / 150 · MHT CET 2023 (11 May Shift 1) · Mathematics

$\int \frac{\mathrm{e}^x(1+x)}{\cos ^2\left(\mathrm{e}^x \cdot x\right)} \mathrm{d} x=$

  • A
    $-\cot \left(e^x\right)+c$, where $c$ is a constant of integration.
  • Correct answer
    B
    $\tan \left(x \cdot \mathrm{e}^x\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • C
    $\tan \left(\mathrm{e}^x\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • D
    $-\cot \left(x \cdot \mathrm{e}^x\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

Let $\mathrm{I}=\int \frac{\mathrm{e}^x(1+x)}{\cos ^2\left(\mathrm{e}^x \cdot x\right)} \mathrm{d} x$ Put $\mathrm{e}^x \cdot x=\mathrm{t} \Rightarrow \mathrm{e}^x(x+1) \mathrm{d} x=\mathrm{dt}$ $$ \begin{aligned} \mathrm{I} & =\int \frac{\mathrm{dt}}{\cos ^2 \mathrm{t}} \\ & =\int \sec ^2 \mathrm{t} d \\ & =\tan \mathrm{t}+\mathrm{c} \\ \therefore \quad \mathrm{I} & =\tan \left(x \cdot \mathrm{e}^x\right)+\mathrm{c} \end{aligned} $$

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

Relevant Questions