Evaluate the following integrals: 2x-x^2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\sqrt{2\text{x}-\text{x}^2}\text{dx}$

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Answer

Let $\text{I}=\int\sqrt{2\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{\text{x}(2-\text{x})}\text{dx}$
Let $\text{x}=1+\sin\text{u}$
or, $\text{dx}=\cos\text{u du}$
$\Rightarrow\text{I}=\int\sqrt{(1+\sin\text{u})(1-\sin\text{u})}\cos\text{u du}$
$\Rightarrow\text{I}=\int\cos^2\text{u du}$
$\Rightarrow\text{I}=\frac{1}{2}\int(\cos2\text{u}+1)\text{du}$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{2}\sin2\text{u}+\text{u}\Big)+\text{C}$
$\Rightarrow\text{I}=\frac{1}{2}(\sin\text{u}\cos\text{u}+\text{u})+\text{C}$
$\Rightarrow\text{I}=\frac{1}{2}\big(\sin\text{u}\sqrt{1-\sin^2\text{u}}+\text{u}\big)+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}(\text{x}-1)\sqrt{2\text{x}-\text{x}^2}+\frac{1}{2}\sin^{-1}(\text{x}-1)+\text{C}$ $\big[\because\text{u}=\sin^{-1}(\text{x}-1)\big]$

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