Integral Calculus Question 232
Question: The value of $ \int{\frac{\sqrt{(x^{2}-a^{2})}}{x}dx} $ will be
[UPSEAT 1999]
Options:
A) $ \sqrt{(x^{2}-a^{2})},-a{{\tan }^{-1}}[ \frac{\sqrt{(x^{2}-a^{2})}}{a} ] $
B) $ \sqrt{(x^{2}-a^{2})},+a{{\tan }^{-1}}[ \frac{\sqrt{(x^{2}-a^{2})}}{a} ] $
C) $ \sqrt{(x^{2}-a^{2})},+a^{2}{{\tan }^{-1}}[\sqrt{x^{2}-a^{2}}] $
D) $ {{\tan }^{-1}}x/a+c $
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Answer:
Correct Answer: A
Solution:
Let $ \sqrt{(x^{2}-a^{2})}=t $
Þ $ x^{2}-a^{2}=t^{2} $
Þ $ x^{2}=a^{2}+t^{2} $
$ \therefore $ $ xdx=tdt $
\ $ \int{\frac{\sqrt{(x^{2}-a^{2})}}{x}dx}=\int{\frac{\sqrt{(x^{2}-a^{2})},x}{x^{2}}dx} $
Þ $ I=\int{\frac{t}{a^{2}+t^{2}}tdt} $ $ =\int{\frac{t^{2}}{a^{2}+t^{2}}dt} $
Þ $ I=\int{( 1-\frac{a^{2}}{a^{2}+t^{2}} ),dt} $ $ =t-a^{2}\frac{1}{a}{{\tan }^{-1}}( \frac{t}{a} ) $
Þ $ I=\sqrt{(x^{2}-a^{2})},-a{{\tan }^{-1}}[ \frac{{ \sqrt{(x^{2}-a^{2})} }}{a} ] $ .
BETA
