Question

Solve $\int \frac{dx}{\sqrt{x-x^2}}dx$ .

• A. $I={\sin }^{-1}(2x-1)+C$
• B. $I={\cos }^{-1}(2x-1)+C$
• C. $I={\sin }^{-1}(x-1)+C$
• D. None of these

Answer 1

Answer: (A)

$I={\sin }^{-1}(2x-1)+C$

Consider the given integral.

$I=\int \frac{dx}{\sqrt{x-x^2}}$

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

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

We know that

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

Therefore,

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

$I={\sin }^{-1}(2x-1)+C$

Hence, this is the answer.