Integral Calculus Question 19
Question: $ \int_{{}}^{{}}{\frac{dx}{\sqrt{x^{2}-a^{2}}}} $ equals
[SCRA 1996]
Options:
A) $ {{\sin }^{-1}}( \frac{x}{a} )+c $
B) $ {\log_{e}}|x+\sqrt{x^{2}-a^{2}}|+c $
C) $ {\log_{e}}|x-\sqrt{x^{2}-a^{2}}|+c $
D) $ \frac{x\sqrt{x^{2}-a^{2}}}{2+c} $
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Answer:
Correct Answer: B
Solution:
$ I=\int_{{}}^{{}}{\frac{dx}{\sqrt{x^{2}-a^{2}}}} $ . Put $ x=a\sec \theta \Rightarrow dx=a\sec \theta ,.,\tan \theta ,d\theta $
$ \therefore ,I=\int_{{}}^{{}}{\frac{a\sec \theta ,.,\tan \theta ,d\theta }{a\tan \theta }}=\int_{{}}^{{}}{\sec \theta ,d\theta } $ $ =\log (\sec \theta +\tan \theta )+=\log ( \frac{x}{a}+\frac{\sqrt{x^{2}-a^{2}}}{a} )+c $ $ =\log (x+\sqrt{x^{2}-a^{2}})+c $ .
BETA
