Mathematics

Solve : $$\displaystyle \int \, \dfrac{dx}{\sqrt{2x - x^2}}$$


SOLUTION
$$I=\displaystyle\int \dfrac{dx}{\sqrt{2x-x^2}}=\displaystyle\int \dfrac{dx}{\sqrt{1-1+2x-x^2}}\cdot \displaystyle\int \dfrac{dx}{\sqrt{1-(x^2-2x+1)}}$$

$$I=\displaystyle\int\dfrac{dx}{\sqrt{1^2-(x-1)^2}}$$

Put $$(x-1)=t$$

$$dx=dt$$.
$$I=\displaystyle\int\dfrac{dt}{\sqrt{1-t^2}}$$

Put $$t=\sin\theta$$

$$dt=\cos\theta d\theta$$

$$I=\displaystyle\int \dfrac{\cos \theta d\theta}{\sqrt{1-\sin^2\theta}}=\displaystyle\int \dfrac{\cos\theta d\theta}{\cos\theta}=\displaystyle\int d\theta$$
$$I=\theta +c=\sin^{-1}(t)+c=\sin^{-1}(x-1)+c$$
$$\therefore \displaystyle\int \dfrac{dx}{\sqrt{2x-x^2}}=\sin^{-1}(x-1)+c$$.
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Subjective Medium Published on 17th 09, 2020
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