Evaluate the following integrals: ^ -1 x dx - Vidyadip

Question

Evaluate the following integrals:

$\int\sin^{-1}\sqrt{\text{x}}\text{dx}$

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Answer

$\int\sin^{-1}\sqrt{\text{x}}\text{dx}$
Let $\text{x} = \text{t}^{2}\ \ \ [\therefore\text{dx = 2tdt}]$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\int\sin^{-1}\sqrt{\text{t}^2}2\text{tdt}=\int\sin^{-1}\text{t}2\text{tdt}$
$=\sin^{-1}\text{t}\int2\text{tdt}-\Big(\int\frac{\text{d}\sin^{-1}\text{t}}{\text{dt}}\big(\int2\text{tdt}\big)\text{dt}\Big)$
$=\sin^{-1}\text{t}(\text{t}^2)-\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}$
Lets solve $\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}$
$\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}=\int\frac{\text{t}^2-1+1}{\sqrt{1-\text{t}^2}}\text{dt}=\int\frac{\text{t}^2-1}{\sqrt{1-\text{t}^2}}\text{dt}+\int\frac{1}{\sqrt{1-\text{t}^2}}\text{dt}$
We know that, value of $\int\frac{1}{\sqrt{1-\text{t}^2}}\text{dt}=\sin^{-1}\text{t}$
Remaining integral to evalute is $\int\frac{\text{t}^2-1}{\sqrt{1-\text{t}^2}}\text{dt}=\int-\sqrt{1-\text{t}^2}\text{dt}$
sub $\text{t}=\sin\text{u},\text{dt}=\cos\text{u du}$
$\int-\sqrt{1-\text{t}^2}\text{dt}=\int-\cos^2\text{u du}=-\int\Big[\frac{1+\cos2\text{u}}{2}\Big]\text{du}$
$=-\frac{\text{u}}{2}-\frac{\sin2\text{u}}{4}$
Substitute back $\text{u}=\sin^{-1}\text{t}$ and $\text{t}=\sqrt{\text{x}}$
$=-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sin(2\sin^{-1}\sqrt{\text{x}})}{4}$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\text{x}\sin^{-1}\sqrt{\text{x}}-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sin(2\sin^{-1}\sqrt{\text{x}})}{4}$
$\sin(2\sin^{-1}\sqrt{\text{x}})=2\sqrt{\text{x}}\sqrt{1-\text{x}}$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\text{x}\sin^{-1}\sqrt{\text{x}}-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sqrt{\text{x}(1-\text{x})}}{2}$

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