Question
Evaluate the following integrals:$\int\frac{\text{e}^\text{x}}{\sqrt{16-\text{e}^{2\text{x}}}}\text{ dx}$

Answer

$\int\frac{\text{e}^\text{x}\text{dx}}{\sqrt{16-(\text{e}^{\text{x}})^2}}$ Let $\text{e}^\text{x}=\text{t}$ $\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{\text{e}^\text{x}\text{dx}}{\sqrt{16-(\text{e}^{\text{x}})^2}}$$=\int\frac{\text{dt}}{\sqrt{16-\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{4^2-\text{t}^2}}$
$=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\text{e}^\text{x}}{4}\Big)+\text{C}$

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