Question

Prove

$$\begin{gathered} \left(a\right)\int \frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}xdx+\mathrm{C}\mathrm{using}\mathrm{differentiation}\mathrm{formulas} \\ \left(b\right)\int \frac{1}{1+x^2}dx={\tan }^{-1}xdx+\mathrm{C}\text{'} \\ \left(c\right)\int \frac{1}{\left|x\right|\sqrt{x^2-1}}dx=se{c}^{-1}xdx+\mathrm{C} \end{gathered}$$

Short answer

Hence Proved

Step-by-step solution

Step 1. To prove

$$\begin{gathered} \left(a\right)\int \frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}xdx+\mathrm{C}\mathrm{using}\mathrm{differentiation}\mathrm{formulas} \\ \left(b\right)\int \frac{1}{1+x^2}dx={\tan }^{-1}xdx+\mathrm{C}\text{'} \\ \left(c\right)\int \frac{1}{\left|x\right|\sqrt{x^2-1}}dx=se{c}^{-1}xdx+\mathrm{C} \end{gathered}$$

Part(a) Step 2. Proving

$$\begin{gathered} \left(a\right)\int \frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}xdx+\mathrm{C}\mathrm{using}\mathrm{differentiation}\mathrm{formulas} \\ \frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}} \\ \mathrm{so},{\sin }^{-1}\mathrm{x}\mathrm{is}\mathrm{the}\mathrm{antiderivative}\mathrm{of}\frac{1}{\sqrt{1-x^2}} \\ \mathrm{Hence}\mathrm{Proved} \end{gathered}$$

Part(b)   Step 3. Proving

$$\begin{gathered} \left(b\right)\int \frac{1}{1+x^2}dx={\tan }^{-1}xdx+\mathrm{C}\mathrm{using}\mathrm{differentiation}\mathrm{formulas} \\ \frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2} \\ \mathrm{so},{\tan }^{-1}\mathrm{x}\mathrm{is}\mathrm{the}\mathrm{antiderivative}\mathrm{of}\frac{1}{1+x^2} \\ \mathrm{Hence}\mathrm{Proved} \end{gathered}$$

Part (c) Step 4. Proving

$$\begin{gathered} \left(c\right)\int \frac{1}{\left|x\right|\sqrt{x^2-1}}dx=se{c}^{-1}xdx+\mathrm{C}\mathrm{using}\mathrm{differentiation}\mathrm{formulas} \\ \frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}} \\ \mathrm{so},{\sec }^{-1}\mathrm{x}\mathrm{is}\mathrm{the}\mathrm{antiderivative}\mathrm{of}\frac{1}{\left|x\right|\sqrt{x^2-1}} \\ \mathrm{Hence}\mathrm{Proved} \end{gathered}$$