Mathematics

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


SOLUTION
Let $$I=\displaystyle\int \sqrt{x-x^2}\ d x$$

$$\implies I=\displaystyle\int \sqrt{-\left(-2\bigg(\dfrac{1}{2}\bigg)(x)+x^2\right)}\ d x$$

$$\implies I=\displaystyle\int \sqrt{\bigg(\dfrac{1}{2}\bigg)^2-\left(\bigg(\dfrac{1}{2}\bigg)^2-2(a)(x)+x^2\right)}\ d x$$

$$\implies I=\displaystyle\int \sqrt{\dfrac{1}{4}-\left(\dfrac{1}{2}-x\right)^2}\ d x$$

Put $$t=a-x\implies d t=-d x$$

$$\implies I=-\displaystyle\int \sqrt{\bigg(\dfrac{1}{2}\bigg)^2-t^2}\ d t$$

As we know that

$$\displaystyle\int \sqrt{a^2-x^2}\ d x=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\text{sin}^{-1} \left(\dfrac{x}{a}\right)+C$$

$$I=-\dfrac{t}{2}\sqrt{\dfrac{1}{4}-t^2}-\dfrac{1/4}{2}\text{sin}^{-1} \left(\dfrac{t}{1/2}\right)+C$$

$$I=-\dfrac{\dfrac{1}{2}-x}{2}\sqrt{\dfrac{1}{4}-\bigg(\dfrac{1}{2}-x\bigg)^2}-\dfrac{1/4}{2}\text{sin}^{-1} \left(2\bigg(x-\dfrac{1}{2}\bigg)\right)+C$$

$$I=\dfrac{1}{4}(2 x-1)\sqrt{x-x^2}-\dfrac{1}{8}\text{sin}^{-1} \left(2 x-1\right)+C$$
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