# Evaluate the integral:

Question:

Evaluate the integral:

$\int \sqrt{x-x^{2}} d x$

Solution:

Key points to solve the problem:

- Such problems require the use of the method of substitution along with a method of integration by parts. By the method of integration by parts if we have $\int \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx}=\mathrm{f}(\mathrm{x}) \int \mathrm{g}(\mathrm{x}) \mathrm{dx}-\int \mathrm{f}^{\prime}(\mathrm{x})\left(\int \mathrm{g}(\mathrm{x}) \mathrm{dx}\right) \mathrm{dx}$

- To solve the integrals of the form: $\int \sqrt{a x^{2}+b x+c} d x$ after applying substitution and integration by parts we have direct formulae as described below:

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

$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$

$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

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

$\therefore I=\int \sqrt{-\left(x^{2}-2\left(\frac{1}{2}\right) x\right)} d x=\int \sqrt{\frac{1}{4}-\left(x^{2}-2\left(\frac{1}{2}\right) x+\left(\frac{1}{2}\right)^{2}\right)} d x$

Using $a^{2}-2 a b+b^{2}=(a-b)^{2}$

We have:

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

As I match with the form: $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$

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

$\Rightarrow I=\frac{1}{4}(2 x-1) \sqrt{x-x^{2}}+\frac{1}{8} \sin ^{-1}(2 x-1)+C$