Question

$\int \frac{1}{\sqrt{7-x^2}}dx=$

• A. $\frac{1}{2\sqrt{7}}\log \left(\frac{\sqrt{7}+x}{\sqrt{7}-x}\right)+C$
• B. ${\sin }^{-1}\left(\frac{x}{\sqrt{7}}\right)+C$
• C. $\log \left|x+{\sqrt{x}}^2-7\right|+C$
• D. $\frac{1}{2\sqrt{7}}\log \left|\frac{x-\sqrt{7}}{x+\sqrt{7}}\right|+C$

Answer 1

The correct option is B

${\sin }^{-1}\left(\frac{x}{\sqrt{7}}\right)+C$

Explanation for the correct answer:

Finding the value for the given integral:

The given equation is$\int \frac{1}{\sqrt{7-x^2}}dx$

Substituting the value of $x=\sqrt{7}\sin u$

$\Rightarrow dx=\sqrt{7}cosudu$

$$\begin{gathered} \int \frac{1}{\sqrt{7-x^2}}dx=\int \frac{1}{\sqrt{7-{(\sqrt{7}\sin u)}^2}}dx \\ =\int \frac{1}{\sqrt{7(1-{\sin }^{2}u)}}dx[\because 1-{\sin }^{2}u={\cos }^{2}u] \\ =\int \frac{1}{\sqrt{7}\sqrt{{\cos }^{2}u}}dx \\ =\int \frac{1}{\sqrt{7}\cos u}dx[\because du=\frac{dx}{\sqrt{7}cosu}] \\ =\int du \\ =u+C \end{gathered}$$

where,

$$\begin{gathered} x=\sqrt{7}\sin u \\ \Rightarrow \sin u=\frac{x}{\sqrt{7}} \\ \Rightarrow u={\sin }^{-1}\frac{x}{\sqrt{7}} \end{gathered}$$

Therefore, $\int \frac{1}{\sqrt{7-x^2}}dx={\sin }^{-1}\frac{x}{\sqrt{7}}+C$

Hence, option (B) is the correct answer.