Question

$\int \frac{dx}{x\sqrt{1-x^2}}$ is equal to

Answer 1

We have,

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

Put $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$

$\begin{array}{c}\int \frac{\cos \theta d\theta }{\sin \theta \cos \theta }\\ =\int \cos ec\theta d\theta \\ =\log \left(\cos ec\theta -\cot \theta \right)+C\\ =\log \left(\frac{1}{x}-\sqrt{{\left(\frac{1}{x}\right)}^2-1}\right)+C\\ =\log \left(\frac{1}{x}-\frac{\sqrt{1-x^2}}{x}\right)+C\\ =\log \left(\frac{1-\sqrt{1-x^2}}{x}\right)+C\end{array}$

Hence, this is the answer.