Question

$\mathbf{\int }\frac{\mathbf{Kdr}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{K}}^{\mathbf{2}}{\mathbf{r}}^{\mathbf{2}}}}\mathbf{=}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{Kr}\mathbf{}\mathbf{or}\mathbf{}\mathbf{-}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{Kr}\mathbf{}\mathbf{or}\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{Kr}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{K}}^{\mathbf{2}}{\mathbf{r}}^{\mathbf{2}}}}$

Hints: Sketch a right triangle with acute angles u and v and label the sides so that sinu=Kr. Also, note that$\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{=}\mathbf{\tau \tau }\mathbf{/}\mathbf{2}$; then if u is an indefinite integral, so is -v since they differ by a constant of integration.

Short answer

The required results are,

$$\begin{gathered} \int \frac{\mathrm{K}}{\sqrt{1-{\mathrm{K}}^2{\mathrm{r}}^2}}\mathrm{dr}={\tan }^{-1}\frac{\mathrm{Kr}}{\sqrt{1-{\mathrm{K}}^2{\mathrm{r}}^2}} \\ ={\sin }^{-1}\mathrm{Kr} \\ =-{\cos }^{-1}\mathrm{Kr} \end{gathered}$$

Step-by-step solution

Definition of Definite Integral

A function that practices the antiderivative of another function is called an indefinite integral. It can be represented graphically as an integral symbol, a function, and finally a dx. The indefinite integral is a more straightforward way of expressing the antiderivative.

Proving the given identity

The following substitution must be used to prove the provided identities,

$\mathrm{r}=\frac{1}{\mathrm{K}}\mathrm{sin\theta }$

By differentiating, we get

$\mathrm{dr}=\frac{1}{\mathrm{K}}\mathrm{cos\theta d\theta }$

Now, replace the above values in the given integration

role="math" localid="1658829206174" $$\begin{gathered} \int \frac{\mathrm{K}}{\sqrt{1-{\mathrm{K}}^1{\mathrm{r}}^2}}\mathrm{d\theta }=\int \frac{1}{\sqrt{1-{\mathrm{K}}^1}{\displaystyle \frac{1}{{\mathrm{K}}^1{\sin }^2\mathrm{e}}}}\times \frac{1}{\mathrm{K}}\mathrm{cos\theta d\theta } \\ =\int \frac{1}{\mathrm{cos\theta }}\mathrm{cos\theta d\theta } \\ =\int 1-\mathrm{d\theta } \\ =\mathrm{\theta } \end{gathered}$$

To find the value of $\theta$the substitute in the above calculation :

$$\begin{gathered} \int \frac{\mathrm{K}}{\sqrt{1-{\mathrm{K}}^2{\mathrm{r}}^3}}\mathrm{dr}=\mathrm{\theta } \\ ={\sin }^{-1}\mathrm{Kr} \end{gathered}$$

The result is ${\sin }^{-1}\mathrm{Kr}$

Another Substitution

To obtain the other outcome, we examine the following substitution:

$\mathrm{r}=\frac{1}{\mathrm{K}}\mathrm{cos\theta }$

By differentiating, we get

$\mathrm{dr}=-\frac{1}{\mathrm{K}}\mathrm{sin\theta d\theta }$

Now, replace the above values in the given integration

$$\begin{gathered} \int \frac{\mathrm{K}}{\sqrt{1-{\mathrm{K}}^2{\mathrm{r}}^2}}\mathrm{dr}=\int \frac{\mathrm{K}}{\sqrt{1-{\mathrm{K}}^2{\displaystyle \frac{1}{{\mathrm{K}}^2}}{\cos }^2\mathrm{\theta }}}\left(\frac{1}{\mathrm{K}}\mathrm{sin\theta d\theta }\right) \\ =-\int 1\times \mathrm{d\theta } \\ =-\mathrm{\theta } \end{gathered}$$

To find the value of the substitute in the above calculation:

$$\begin{gathered} \int \frac{\mathrm{K}}{\sqrt{1-{\mathrm{K}}^2{\mathrm{r}}^2}}\mathrm{dr}=-\mathrm{\theta } \\ =-{\cos }^{-1}\mathrm{Kr} \end{gathered}$$

Finding the final result

Consider the trigonometric identity

$\tan \theta =\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}$

By substituting Kr=sin$\theta$

Now, if we apply the above substitution, we get

$$\begin{gathered} \mathrm{tan\theta }=\frac{\mathrm{sin\theta }}{\sqrt{1-{\sin }^2\mathrm{\theta }}} \\ \mathrm{tan\theta }=\frac{\mathrm{Kr}}{\sqrt{1-{\mathrm{K}}^2{\mathrm{r}}^2}} \\ \mathrm{\theta }={\tan }^{-1}\frac{\mathrm{Kr}}{\sqrt{1-{\mathrm{K}}^2{\mathrm{r}}^2}} \end{gathered}$$

So, the required solution is :$\int \frac{Kr}{\sqrt{1-K^2{r}^2}}\mathrm{dr}={\tan }^{-1}\frac{\mathrm{Kr}}{\sqrt{1-{\mathrm{K}}^2{\mathrm{r}}^2}}$