Question

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Integrate the integral:

$\int \dfrac{1}{\sqrt{x} \sqrt{1-x} }$​

Answer 1

We need to find,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\sqrt{x}\sqrt{1-x}}$

or,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\sqrt{x-x^2}}$

Pre-adding and subtracting $\dfrac{1}{4}$ in the denominator surd,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\sqrt{\dfrac{1}{4}-\dfrac{1}{4}+x-x^2}}$

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

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

or,

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

We have,

• $\displaystyle\int\dfrac{dx}{\sqrt{c^2-(ax+b)^2}}=\dfrac{1}{a}\sin^{-1}\left(\dfrac{ax+b}{c}\right)+C$

Here,

• $a=-1$

• $b=\dfrac{1}{2}$

• $c=\dfrac{1}{2}$

Hence,

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

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

Since $\sin^{-1}(-x)=-\sin^{-1}x,$

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