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$\int \frac{d x}{\sqrt{x-x^2}}$ is equal to
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The correct answer is:
$2 \sin ^{-1} \sqrt{x}+C$
Let $\begin{aligned} I & =\int \frac{d x}{\sqrt{x-x^2}} \\ & =\int \frac{1}{\sqrt{x}} \times \frac{d x}{\sqrt{1-x}}\end{aligned}$
Put $\quad \sqrt{x}=\sin \theta$
$\begin{aligned} \Rightarrow \quad \frac{1}{2 \sqrt{x}} d x & =\cos \theta d \theta \\ \therefore \quad I & =\int \frac{2 \cos \theta d \theta}{\sqrt{1-\sin ^2 \theta}} \\ & =\int \frac{2 \cos \theta}{\cos \theta} d \theta \\ & =\int 2 d \theta=2 \theta+C \\ & =2 \sin ^{-1} \sqrt{x}+C\end{aligned}$
Put $\quad \sqrt{x}=\sin \theta$
$\begin{aligned} \Rightarrow \quad \frac{1}{2 \sqrt{x}} d x & =\cos \theta d \theta \\ \therefore \quad I & =\int \frac{2 \cos \theta d \theta}{\sqrt{1-\sin ^2 \theta}} \\ & =\int \frac{2 \cos \theta}{\cos \theta} d \theta \\ & =\int 2 d \theta=2 \theta+C \\ & =2 \sin ^{-1} \sqrt{x}+C\end{aligned}$
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