Question

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

• A. $\sin \left({\tan }^{-1}x\right)+c$
• B. $\tan \left({\text{sec}}^{-1}x\right)+c$
• C. $\tan \left({\sin }^{-1}x\right)+c$
• D. $-\tan \left({\cos }^{-1}x\right)+c$

Answer 1

The correct option is C

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

Explanation for correct option:

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

Consider the given equation as

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

Let us assume that,

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

Substitute the above values in the integral becomes

$$\begin{gathered} I=\int se{c}^2\left(t\right)dt \\ I=\tan t+c \\ I=\tan \left({\sin }^{-1}x\right)+c\left[where,t={\sin }^{-1}x\right] \end{gathered}$$

Hence, the correct answer is option C.