Integral Calculus Question 2
Question: $ \int_{{}}^{{}}{\frac{{{\sec }^{2}}x\ dx}{\sqrt{{{\tan }^{2}}x+4}}=} $
Options:
A) $ \log [ \tan x+\sqrt{{{\tan }^{2}}x+4} ]+c $
B) $ \frac{1}{2}\log [ \tan x+\sqrt{{{\tan }^{2}}x+4} ]+c $
C) $ \log [ \frac{1}{2}\tan x+\frac{1}{2}\sqrt{{{\tan }^{2}}x+4} ]+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ t=\tan x\Rightarrow dt={{\sec }^{2}}x,dx, $ then $ \int_{{}}^{{}}{\frac{{{\sec }^{2}}x,dx}{\sqrt{{{\tan }^{2}}x+4}}}=\int_{{}}^{{}}{\frac{1}{\sqrt{t^{2}+2^{2}}}},dt $ $ =\log [\tan x+\sqrt{{{\tan }^{2}}x+4}]+c. $
BETA
