Q. 68 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}\mathrm{d}x=\frac{1}{2}{\left({\sin }^{-1}\mathrm{x}\right)}^2+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals.

$\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}\mathrm{d}x$

Step 2. Using the substitution method.

$$\begin{gathered} u={\sin }^{-1}x \\ \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}} \\ du=\frac{1}{\sqrt{1-x^2}}dx \end{gathered}$$

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}\mathrm{d}x=\int u\mathrm{du} \\ \int \frac{{\sin }^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}=\left[\frac{{\mathrm{u}}^{1+1}}{1+1}\right]+\mathrm{C} \\ \int \frac{{\sin }^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}=\frac{{\mathrm{u}}^2}{2}+\mathrm{C} \\ \int \frac{{\sin }^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}=\frac{1}{2}{\mathrm{u}}^2+\mathrm{C} \\ \int \frac{{\sin }^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}=\frac{1}{2}{\left({\sin }^{-1}\mathrm{x}\right)}^2+\mathrm{C} \end{gathered}$$