Question
Evaluate the following integrals:
$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
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Answer
$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{-(\text{x}^2-\text{x})}\text{dx}$
$=\int\sqrt{-\bigg\{\text{x}^2-\text{x}+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}\text{dx}$
$=\int\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$=\Big(\frac{\text{x}-\frac{1}{2}}{2}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\bigg)+\text{C}$
$\Big[\because\ \int\sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{a}^2-\text{x}^2}+\frac{1}{2}\text{a}^2\sin^{-1}\frac{\text{x}}{\text{a}}=\text{C}\Big]$
$=\Big(\frac{2\text{x}-1}{4}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}(2\text{x}-1)+\text{C}$
$=\int\sqrt{-(\text{x}^2-\text{x})}\text{dx}$
$=\int\sqrt{-\bigg\{\text{x}^2-\text{x}+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}\text{dx}$
$=\int\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$=\Big(\frac{\text{x}-\frac{1}{2}}{2}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\bigg)+\text{C}$
$\Big[\because\ \int\sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{a}^2-\text{x}^2}+\frac{1}{2}\text{a}^2\sin^{-1}\frac{\text{x}}{\text{a}}=\text{C}\Big]$
$=\Big(\frac{2\text{x}-1}{4}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}(2\text{x}-1)+\text{C}$
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