$\int \frac{\mathrm{e}^x(1+x)}{\cos ^2\left(\mathrm{e}^x \cdot x\right)} \mathrm{d} x=$
$\int \frac{\mathrm{e}^x(1+x)}{\cos ^2\left(\mathrm{e}^x \cdot x\right)} \mathrm{d} x=$
- $-\cot \left(e^x\right)+c$, where $c$ is a constant of integration.
- $\tan \left(x \cdot \mathrm{e}^x\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\tan \left(\mathrm{e}^x\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $-\cot \left(x \cdot \mathrm{e}^x\right)+\mathrm{c}$, 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 $\mathrm{e}^x \cdot x=\mathrm{t} \Rightarrow \mathrm{e}^x(x+1) \mathrm{d} x=\mathrm{dt}$
$\begin{aligned}
\mathrm{I} & =\int \frac{\mathrm{dt}}{\cos ^2 \mathrm{t}} \\
& =\int \sec ^2 \mathrm{t} dt \\
& =\tan \mathrm{t}+\mathrm{c} \\
\therefore \quad \mathrm{I} & =\tan \left(x \cdot \mathrm{e}^x\right)+\mathrm{c}
\end{aligned}$
Asked in: MHT CET 2023 (11 May Shift 1)
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