Integral Calculus Question 472
Question: $ \int{\frac{dx}{{{(a^{2}+x^{2})}^{3/2}}}} $ is equal to
[RPET 2000]
Options:
A) $ \frac{x}{{{( a^{2}+x^{2} )}^{1/2}}} $
B) $ \frac{x}{a^{2}{{( a^{2}+x^{2} )}^{1/2}}} $
C) $ \frac{1}{a^{2}{{( a^{2}+x^{2} )}^{1/2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\frac{1}{2}\log | \frac{1+\tan x}{1-\tan x} |+C $ Put $ x=a\tan \theta \Rightarrow dx=a{{\sec }^{2}}\theta ,d\theta $
$ \therefore ,I=\int{\frac{a{{\sec }^{2}}\theta }{{{(a^{2}+a^{2}{{\tan }^{2}}\theta )}^{3/2}}}d\theta }=\int{\frac{a{{\sec }^{2}}\theta }{a^{3}{{({{\sec }^{2}}\theta )}^{3/2}}}d\theta } $
Þ $ I=\frac{1}{a^{2}}\int{\frac{d\theta }{\sec \theta }} $ $ =\frac{1}{a^{2}}\int{\cos \theta ,d\theta =\frac{1}{a^{2}}\sin \theta +c} $
Þ $ I=\frac{x}{a^{2}{{(x^{2}+a^{2})}^{1/2}}}+c $ .
BETA
