Q. 39 - Calculus

Question

Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess-and-check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. 

$\int \frac{7}{\sqrt{1-{\left(2x\right)}^2}}dx$

Step-by-Step Solution

Answer

The solution of the integral is $\frac{7}{2}{\sin }^{-1}\left(2x\right)+C.$

Step 1. Given Information.

The given integral is$\int \frac{7}{\sqrt{1-{\left(2x\right)}^2}}dx.$

Step 2. Solve.

By solving the integral we get, 

$$\begin{gathered} \int \frac{7}{\sqrt{1-{\left(2x\right)}^2}}dx \\ \mathrm{Let} t=2x, dt=2dx \\ =7\int \frac{1}{\sqrt{1-{\left(t\right)}^2}}\left(\frac{dt}{2}\right) \\ =\frac{7}{2}\int \frac{dt}{\sqrt{1-{\left(t\right)}^2}} \\ =\frac{7}{2}{\sin }^{-1}\left(t\right)+C \\ \mathrm{Substiute} \mathrm{back} t=2x \\ =\frac{7}{2}{\sin }^{-1}\left(2x\right)+C \end{gathered}$$

Step 3. Verification.

To verify the answer we differentiate$\frac{7}{2}{\sin }^{-1}\left(2x\right)+C$  it.

On differentiating we get,

$$\begin{gathered} \frac{7}{2}{\sin }^{-1}\left(2x\right)+C \\ =\frac{d}{dx}\left(\frac{7}{2}{\sin }^{-1}\left(2x\right)\right)+\frac{d}{dx}C \\ =\frac{7}{\sqrt{1-{\left(2x\right)}^2}} \end{gathered}$$

Hence proved.