Evaluate the following integrals: ^ -1 x dx

Question

Evaluate the following integrals:$\int\sec^{-1}\sqrt{\text{x}}\text{dx}$

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Answer

$\int1.\sec^{-1}\sqrt{\text{x}}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\Big(\sec^{-1}\sqrt{\text{x}}\Big)\int1\text{dx}\Big\}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}.\text{x}-\int\frac{1}{\sqrt{\text{x}}\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\times\text{x dx}$
$=\text{x}\sec^{-1}\sqrt{\text{x}}-\frac{1}{2}\int(1-\text{x})^{-\frac{1}{2}\text{dx}}$
$=\text{x}\sec^{-1}\text{x}-\frac{1}{2}\Bigg[\frac{(1-\text{x})^{-\frac{1}{2}+1}}{\big(-\frac{1}{2}+1\big)(-1)}\Bigg]+\text{C}$
$=\text{x}\sec^{-1}\text{x}+(1-\text{x})^{\frac{1}{2}}+\text{C}$

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