Question

integral DX by root x minus x square ​

Answer 1

Answer:

Step-by-step:

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Answer 2

We need to recall the following integration formula.

• $\int\frac{1}{\sqrt{1-x^2} }dx=sin^{-1}x+C$

This problem is about integration.

Given:

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

Rewrite the denominator as a complete square.

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

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

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

Substitute $u=2x-1$.

Differentiate both sides, we get

$du=2dx$

$\frac{du}{2}=dx$

Then,

$I=\int\frac{1}{2\sqrt{\frac{1}{4}-\frac{u^2}{4} } }du$

$I=\int\frac{1}{2\sqrt{\frac{1-u^2}{4} } }du$

$I=\int\frac{1}{\frac{2}{2} \sqrt{{1-u^2} } }du$

$I=\int\frac{1}{\sqrt{{1-u^2} } }du$

$I=sin^{-1}u+C$

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

Thus, $\int\frac{dx}{\sqrt{x-x^2} }=sin^{-1}(2x-1)+C$