Q. 75 - Calculus

Question

Prove 

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

Step-by-Step Solution

Answer

Hence Proved

Step 1. To prove

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

Part(a) Step 2. Proving

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

Part(b) Step 3. Proving

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

Part (c) Step 4. Proving

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