A) \[\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]
B) \[\frac{\pi }{4}-{{\cos }^{-1}}{{x}^{2}}\]
C) \[\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]
D) \[\frac{\pi }{4}+{{\cos }^{-1}}{{x}^{2}}\]
Correct Answer: A
Solution :
\[{{x}^{2}}=\cos 2\theta ;\theta =\frac{1}{2}\cos {{x}^{2}}\] \[{{\tan }^{-1}}\left[ \frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }} \right]\] \[{{\tan }^{-1}}\left[ \frac{1+\tan \theta }{1-\tan \theta } \right]\] \[={{\tan }^{-1}}\left[ \tan \left( \frac{\pi }{4}+\theta \right) \right]\] \[=\frac{\pi }{4}+\frac{1}{2}\cos {{x}^{2}}\]
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