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Question

Mathematics Question on Integrals of Some Particular Functions

Integrate sec⁡2 (sin⁡−1x)1−x2\frac{{{\sec }^{2}}\,({{\sin }^{-1}}x)}{\sqrt{1-{{x}^{2}}}}

A

sin⁡ (tan⁡−1x)+c \sin \,({{\tan }^{-1}}x)+c

B

tan⁡ (sec⁡−1x)+c\tan \,({{\sec }^{-1}}x)+c

C

tan⁡ (sin⁡−1x)+c\tan \,({{\sin }^{-1}}x)+c

D

−tan⁡ (cos−1x)+c-\tan \,(co{{s}^{-1}}x)+c

Answer

tan⁡ (sin⁡−1x)+c\tan \,({{\sin }^{-1}}x)+c

Explanation

Solution

Let l=∫sec⁡2(sin⁡−1x)1−x2dxl=\int{\frac{{{\sec }^{2}}({{\sin }^{-1}}x)}{\sqrt{1-{{x}^{2}}}}}dx Again, let sin⁡−1x=t{{\sin }^{-1}}x=t
⇒\Rightarrow dtdx=11−x2\frac{dt}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}}
⇒\Rightarrow dt=11−x2 dxdt=\frac{1}{\sqrt{1-{{x}^{2}}}}\,dx
∴\therefore l=∫sec⁡2 t dtl=\int{{{\sec }^{2}}\,t\,dt}
=tan⁡ t+C=\tan \,t+C
=tan⁡(sin⁡−1x)+C=\tan ({{\sin }^{-1}}x)+C