Question

Fill in the blanks to complete each of the following integration formulas.

$\int {\sin }^{-1}xdx=\_\_\_\_\_\_\_$

Short answer

The result is$x{\sin }^{-1}x+\sqrt{1-x^2}+C$.

Step-by-step solution

Step 1. Given information.

Consider the given integration.

$\int {\sin }^{-1}xdx$

Step 2. Find the integration.

Apply the by-parts integration method. Consider the values as $u=1$and $v={\sin }^{-1}x$.

$\begin{array}{rcl}\int {\sin }^{-1}xdx& =& x{\sin }^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx\\ & =& x{\sin }^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx\left[t=1-x^2,-\frac{dt}{2}=xdx\right]\\ & =& x{\sin }^{-1}x-\left[-2\int \frac{1}{\sqrt{t}}dt\right]\\ & =& x{\sin }^{-1}x-\left[\frac{-2\sqrt{t}}{2}\right]\\ & =& x{\sin }^{-1}x+\sqrt{1-x^2}+C\end{array}$