Evaluate the following integrals: ^ x . 1-x^2 ^ -1 x+1 1-x^2 dx - Vidyadip

Question

Evaluate the following integrals:

$\int\text{e}^{\text{x}}.\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\text{dx}$

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Answer

Let $\text{I}=\int\text{e}^{\text{x}}\bigg[\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\bigg]\text{dx}$
$=\int\text{e}^{\text{x}}\Big[\sin^{-1}\text{x}+\frac{1}{\sqrt{1-\text{x}^2}}\Big]\text{dx}$
Here, $\text{f(x)}=\sin^{-1}\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\sqrt{1-\text{x}^2}}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\sin^{-1}\text{x}=\text{t}$
Diff both sides w.r.t x
$\Big(\text{e}^{\text{x}}\sin^{-1}\text{x}+\text{e}^{\text{x}}\times\frac{1}{\sqrt{1-\text{x}^2}}\Big)\text{dx = dt}$
$\because\text{I}=\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\sin^{-1}\text{x + C}$

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