A) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{2}{x}\]
C) \[-\frac{2}{x}\]
D) \[\frac{-1}{\sqrt{1-{{x}^{2}}}}\]
Correct Answer: C
Solution :
Let \[u=\text{cose}{{\text{c}}^{-1}}\left( \frac{1}{2x\sqrt{1-{{x}^{2}}}} \right)\] and \[v=\sqrt{1-{{x}^{2}}}\] Put \[x=\sin \theta ,\] Then, \[u=\text{cose}{{\text{c}}^{-1}}\left\{ \frac{1}{2\sin \theta .\cos \theta } \right\}=\text{cose}{{\text{c}}^{-1}}\,(\text{cosec}\,\text{ 2}\theta \text{)}\] \[u=2\theta =2{{\sin }^{-1}}x\] ?(i) and \[v=\sqrt{1-{{\sin }^{2}}\theta }\,=\,\cos \theta =\sqrt{1-{{x}^{2}}}\] ?.(ii) Now, \[\frac{du}{dx}=\frac{2}{\sqrt{1-{{x}^{2}}}}\] and \[\frac{dv}{dx}=\frac{-2x}{2\sqrt{1-{{x}^{2}}}}=\frac{-x}{\sqrt{1-{{x}^{2}}}}\]0 \[\Rightarrow \] \[\frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{2/\sqrt{1-{{x}^{2}}}}{-x/\sqrt{1-{{x}^{2}}}}=\frac{-2}{x}\]
You need to login to perform this action.
You will be redirected in
3 sec
