Integral Calculus Question 424
Question: $ \int_{{}}^{{}}{\sec x{{\tan }^{3}}x\ dx=} $
Options:
A) $ \frac{1}{3}{{\sec }^{3}}x-\sec x+c $
B) $ {{\sec }^{3}}x-\sec x+c $
C) $ \frac{1}{3}{{\sec }^{3}}x+\sec x+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\sec x{{\tan }^{3}}x\ dx=\int_{{}}^{{}}{\sec x({{\sec }^{2}}x-1)\tan x\ dx}} $ $ =\int_{{}}^{{}}{\sec x\tan x{{\sec }^{2}}x\ dx}-\int_{{}}^{{}}{\sec x\tan x\ dx} $ $ =\frac{{{\sec }^{3}}x}{3}-\sec x+c $ , (Putting $ \sec x=t $ in first part).
BETA
