Integral Calculus Question 345
Question: $ \int_{{}}^{{}}{\frac{dx}{4{{\sin }^{2}}x+5{{\cos }^{2}}x}=} $
[AISSE 1986]
Options:
A) $ \frac{1}{\sqrt{5}}{{\tan }^{-1}}( \frac{2\tan x}{\sqrt{5}} )+c $
B) $ \frac{1}{\sqrt{5}}{{\tan }^{-1}}( \frac{\tan x}{\sqrt{5}} )+c $
C) $ \frac{1}{2\sqrt{5}}{{\tan }^{-1}}( \frac{2\tan x}{\sqrt{5}} )+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{dx}{4{{\sin }^{2}}x+5{{\cos }^{2}}x}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x,dx}{4{{\tan }^{2}}x+5}=\frac{1}{4}\int_{{}}^{{}}{\frac{{{\sec }^{2}}x,dx}{{{\tan }^{2}}x+\frac{5}{4}}}} $
Put $ \tan x=t\Rightarrow {{\sec }^{2}}x,dx=dt, $ then it reduces to
$ \frac{1}{4}\int_{{}}^{{}}{\frac{dt}{t^{2}+{{( \frac{\sqrt{5}}{2} )}^{2}}}=\frac{2}{4\sqrt{5}}{{\tan }^{-1}}( \frac{2t}{\sqrt{5}} )}+c $
$ =\frac{1}{2\sqrt{5}}{{\tan }^{-1}}( \frac{2\tan x}{\sqrt{5}} )+c. $
BETA
