Integral Calculus Question 417
Question: The value of $ \int{{{\sec }^{3}}xdx} $ will be
[UPSEAT 1999]
Options:
A) $ \frac{1}{2}[ ,\sec x\tan x+\log (\sec x+\tan x) ] $
B) $ \frac{1}{3}[ ,\sec x\tan x+\log (\sec x+\tan x) ] $
C) $ \frac{1}{4}[ ,\sec x\tan x+\log (\sec x+\tan x) ] $
D) $ \frac{1}{8}[ ,\sec x\tan x+\log (\sec x+\tan x) ] $
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Answer:
Correct Answer: A
Solution:
Let $ I=\int{{{\sec }^{3}}xdx} $ $ =\int{\sec x{{\sec }^{2}}xdx} $
$ \Rightarrow I=\sec x\tan x-\int{\sec x,{{\tan }^{2}}x,dx} $
$ \Rightarrow I=\sec x\tan x-\int{\sec x,({{\sec }^{2}}x-1)dx} $
$ \Rightarrow I=\sec x\tan x-\int{{{\sec }^{3}}x,dx}+\int{\sec x,dx} $
$ \Rightarrow \ I=\sec x\tan x-I+\log ,(\sec x,+\tan x,) $
$ \Rightarrow 2I=\sec x\tan x+\log (\sec x+\tan x) $
$ \Rightarrow I=\frac{1}{2}[\sec x\tan x+\log (\sec x+\tan x)] $ .
BETA
