Question

Solve the definite integration.

${\int }_{-\frac{1}{2}}^{\frac{1}{2}}{\sin }^{-1}xdx$

Short answer

The solution is$0$.

Step-by-step solution

Step 1. Given information.

The given integral is${\int }_{-\frac{1}{2}}^{\frac{1}{2}}{\sin }^{-1}xdx$.

Step 2. First, solve the indefinite integral.

$\begin{array}{rcl}I& =& \int {\sin }^{-1}xdx\\ & =& {\sin }^{-1}x\int dx-\int \left[\frac{d}{dx}\left({\sin }^{-1}x\right)\int dx\right]dx\\ & =& {\sin }^{-1}x·x-\int \frac{1}{\sqrt{1-x^2}}·xdx\\ & =& {\sin }^{-1}x·x-\int \frac{x}{\sqrt{1-x^2}}dx\\ & =& x{\sin }^{-1}x-I_1\end{array}$

Step 3. Solve for I1.

$I_1=\int \frac{x}{\sqrt{1-x^2}}dx$

Let $1-x^2=t$, $-2xdx=dt\Rightarrow xdx=-\frac{1}{2}dt$.

$\begin{array}{rcl}{I}_1& =& \int -\frac{1}{2\sqrt{t}}dt\\ & =& -\frac{1}{2}\int t^{-\frac{1}{2}}dt\\ & =& -\frac{1}{2}\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\\ & =& -\frac{1}{2}\frac{t^{{\displaystyle \frac{1}{2}}}}{{\displaystyle \frac{1}{2}}}\\ & =& -\frac{1}{2}2\sqrt{t}\\ & =& -\sqrt{t}\\ & =& -\sqrt{1-x^2}\end{array}$

Step 4. Apply the limit of integration.

$\begin{array}{rcl}{\int }_{-\frac{1}{2}}^{\frac{1}{2}}{\sin }^{-1}x& =& {\left[x{\sin }^{-1}x+\sqrt{1-x^2}\right]}_{-\frac{1}{2}}^{\frac{1}{2}}\\ & =& \left[\frac{1}{2}{\sin }^{-1}\frac{1}{2}+\sqrt{1-{\left(\frac{1}{2}\right)}^2}\right]-\left[-\frac{1}{2}{\sin }^{-1}\left(-\frac{1}{2}\right)+\sqrt{1-{\left(-\frac{1}{2}\right)}^2}\right]\\ & =& \left[\frac{1}{2}{\sin }^{-1}\frac{1}{2}+\sqrt{1-\frac{1}{4}}\right]-\left[-\frac{1}{2}{\sin }^{-1}\left(-\frac{1}{2}\right)+\sqrt{1-\frac{1}{4}}\right]\\ & =& \left[\frac{1}{2}{\sin }^{-1}\frac{1}{2}+\sqrt{\frac{3}{4}}\right]-\left[-\frac{1}{2}{\sin }^{-1}\left(-\frac{1}{2}\right)+\sqrt{\frac{3}{4}}\right]\\ & =& \frac{\mathrm{\pi }}{12}+\frac{\sqrt{3}}{2}-\frac{\mathrm{\pi }}{12}-\frac{\sqrt{3}}{2}\\ & =& 0\end{array}$

Step 5. Simplified answer

Hence, the required value is$0$.