Definite Integration Question 69
Question: $ \frac{1}{2}(e-3) $
[SCRA 1986; Karnataka CET 1999]
Options:
A) $ \pi ab $
B) $ {{\pi }^{2}}ab $
C) $ \frac{\pi }{ab} $
D) $ \frac{\pi }{2ab} $
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Answer:
Correct Answer: D
Solution:
$ I=\int_0^{\pi /2}{\frac{dx}{a^{2}{{\cos }^{2}}x+b^{2}{{\sin }^{2}}x}.} $
Dividing the numerator and denominator by $ {{\cos }^{2}}x, $ we get $ I=\int_0^{\pi /2}{\frac{\frac{1}{{{\cos }^{2}}x}dx}{a^{2}+b^{2}\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}=\int_0^{\pi /2}{\frac{{{\sec }^{2}}x}{a^{2}+b^{2}{{\tan }^{2}}x}dx}} $ . Substituting $ b\tan x=t $ and $ b{{\sec }^{2}}xdx=dt $ and limit when $ x=0 $ , then $ t=0 $ and when $ x=\frac{\pi }{2}, $ then $ t=\infty , $
therefore, $ I=\int_0^{\infty }{\frac{\frac{dt}{b}}{a^{2}+t^{2}}}=\frac{1}{b}[ \frac{1}{a}{{\tan }^{-1}}( \frac{t}{a} ) ]_0^{\infty } $
$ =\frac{1}{ab}[ {{\tan }^{-1}}\infty -{{\tan }^{-1}}0 ]=\frac{1}{ab}( \frac{\pi }{2}-0 )=\frac{\pi }{2ab} $ .
BETA
