Integral Calculus Question 54
Question: $ \int_{{}}^{{}}{\frac{1}{1+{{\cos }^{2}}x}dx}= $
Options:
A) $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}(\tan x)+c $
B) $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{1}{2}\tan x )+c $
C) $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{1}{\sqrt{2}}\tan x )+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{dx}{1+{{\cos }^{2}}x}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x,dx}{{{\sec }^{2}}x+1}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x}{{{\tan }^{2}}x+2}},dx $ $ =\int_{{}}^{{}}{\frac{dt}{t^{2}+2}=\frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{t}{\sqrt{2}} )+c} $ {Putting $ \tan x=t} $ $ =\frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{1}{\sqrt{2}}\tan x )+c $ . Trick : By inspection, $ \frac{d}{dx}{ \frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{1}{\sqrt{2}}\tan x ) }=\frac{1}{\sqrt{2}}( \frac{1}{1+\frac{{{\tan }^{2}}x}{2}} )\frac{1}{\sqrt{2}}{{\sec }^{2}}x $ $ =\frac{1}{2},.,\frac{2{{\sec }^{2}}x}{(2+{{\tan }^{2}}x)}=\frac{{{\sec }^{2}}x}{1+{{\sec }^{2}}x}=\frac{1}{1+{{\cos }^{2}}x} $ .
BETA
