Question

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

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

Answer 1

The correct option is B

$\left(\frac{1}{2}\right){\left({\sin }^{-1}x\right)}^2+c$

Explanation for the correct option:

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

consider the given Equation as

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

Let us assume that,

$$\begin{gathered} t={\sin }^{-1}x \\ dt=\frac{1}{\sqrt{1-x^2}}dx \end{gathered}$$

Then the integral becomes,

$$\begin{gathered} I=\int tdt \\ \Rightarrow I=\frac{t^2}{2}+c \\ \Rightarrow I=\left(\frac{1}{2}\right){\left({\sin }^{-1}x\right)}^2+c\left[\text{where},t={\sin }^{-1}x\right] \end{gathered}$$

Hence, the correct answer is Option B.