Question
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}$

Answer

Let $\text{I}=\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}\ ....(1)$

Let $\text{m}\sin^{-1}\text{x}=\text{t}$ then,

$\text{d}\big(\text{m}\sin^{-1}\text{x}\big)=\text{dt}$

$\Rightarrow\text{m}\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$

$\Rightarrow\frac{\text{dx}}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{m}}$

Putting, $\text{m}\sin^{-1}\text{x}=\text{t}$ and $\frac{\text{dx}}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{m}}$ in equation (1),

We get,

$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{\text{m}}$

$=\frac{1}{\text{m}}\text{e}^\text{t}+\text{C}$

$=\frac{1}{\text{m}}\text{e}^{\text{m}\sin^{-1}\text{x}}+\text{C}$

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