Integral Calculus Question 90
Question: $ \int_{{}}^{{}}{{{\sec }^{4}}x\tan x\ dx=} $
[AI CBSE 1980, 81; SCRA 1996]
Options:
A) $ \frac{1}{4}{{\sec }^{4}}x+c $
B) $ 4{{\sec }^{4}}x+c $
C) $ \frac{{{\sec }^{3}}x}{3}+c $
D) $ 3{{\sec }^{3}}x+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{{{\sec }^{4}}tnx,dx}=\int_{{}}^{{}}{{{\sec }^{3}}x\sec x\tan x,dx} $ Put $ t=\sec x\Rightarrow dt=\sec x\tan x,dx, $ then it reduces to $ \int_{{}}^{{}}{t^{3}dt}=\frac{t^{4}}{4}+c=\frac{1}{4}{{\sec }^{4}}x+c. $
BETA
