Question

Solve each of the integrals in Exercises 39–74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$\int \frac{\sqrt{4-x^2}}{x^2}dx$

Short answer

The solution of the integral is$-\frac{\sqrt{4-x^2}}{x}-{\sin }^{-1}\frac{x}{2}+C.$

Step-by-step solution

Step 1. Given Information.

The given integral is$\int \frac{\sqrt{4-x^2}}{x^2}dx.$

Step 2. Solve.

To solve the integral, let $x=2\sin u,$so the derivation of u is$dx=2\cos udu.$

Thus, substitute u into the original integral.

localid="1648797152038" $$\begin{gathered} \int \frac{\sqrt{4-x^2}}{x^2}dx \\ =\int \frac{\sqrt{4-{\left(2\sin u\right)}^2}}{{\left(2\sin u\right)}^2}2\cos udu \\ =\int \frac{\sqrt{4-\left(4{\sin }^{2}u\right)}}{4{\sin }^{2}u}2\cos udu \\ =\int \frac{\sqrt{4(1-{\sin }^{2}u)}}{4{\sin }^{2}u}2\cos udu \\ =\int \frac{2\sqrt{1-{\sin }^{2}u}}{4{\sin }^{2}u}2\cos udu \\ =\int \frac{\sqrt{1-{\sin }^{2}u}}{{\sin }^{2}u}\cos udu \\ \mathrm{Let}\text{'}\mathrm{s}\mathrm{use}\mathrm{the}\mathrm{identity}si{n}^{2}x+co{s}^{2}x=1 \\ =\int \frac{\sqrt{{\cos }^{2}u}}{{\sin }^{2}u}\cos udu \\ =\int \frac{{\cos }^{2}u}{{\sin }^{2}u}du \end{gathered}$$

Step 3. Solve.

By proceeding with the calculation further,

$$\begin{gathered} =\int \frac{{\cos }^{2}u}{{\sin }^{2}u}du \\ =\int co{t}^{2}udu \\ =\int cs{c}^{2}u-1du \\ =\int cs{c}^{2}udu-\int 1du \\ =-cotu-u+C \end{gathered}$$

Now, substitute back uin the above equation,

$$\begin{gathered} =-cot\left({\sin }^{-1}\left(\frac{x}{2}\right)\right)-{\sin }^{-1}\frac{x}{2}+C \\ \mathrm{Use}\mathrm{the}\mathrm{identity}cot\left({\sin }^{-1}\left(x\right)\right)=\frac{\sqrt{1-x^2}}{x} \\ =-\frac{\sqrt{1-{\left({\displaystyle \frac{x}{2}}\right)}^2}}{{\displaystyle \frac{x}{2}}}-{\sin }^{-1}\frac{x}{2}+C \\ =-\frac{\sqrt{4-x^2}}{x}-{\sin }^{-1}\frac{x}{2}+C \end{gathered}$$