Chapter 5.5, Problem 27E

To determine

To evaluate: The indefinite integral.

Answer to Problem 27E

The integration of the indefinite integral ${\displaystyle \int \frac{dx}{\sqrt{1-x^2}{\sin }^{-1}x}}$is $\ln \left|{\sin }^{-1}x\right|+C$ .

Explanation of Solution

Given information: ${\displaystyle \int \frac{dx}{\sqrt{1-x^2}{\sin }^{-1}x}}$

Calculation: ${\displaystyle \int \frac{dx}{\sqrt{1-x^2}{\sin }^{-1}x}}$

Let apply the substitution $z={\sin }^{-1}x$

Partially differentiate both side,

  $\Rightarrow dz=\frac{1}{\sqrt{1-x^2}}dx$

Let’s evaluate the indefinite integral,

  $={\displaystyle \int \frac{1}{z}dz}$

Therefore, using the common integral,

  $=\ln \left|z\right|$

Now, Substitute back $z={\sin }^{-1}x$ ,

  $=\ln \left|{\sin }^{-1}x\right|$

Adding a constant with the solution,

  $=\ln \left|{\sin }^{-1}x\right|+C$

Hence, The integration of the indefinite integral ${\displaystyle \int \frac{dx}{\sqrt{1-x^2}{\sin }^{-1}x}}$is $\ln \left|{\sin }^{-1}x\right|+C$ .