Integral Calculus Question 219
Question: $ \int_{{}}^{{}}{{{\sec }^{p}}x\tan x\ dx=} $
Options:
A) $ \frac{{{\sec }^{p+1}}x}{p+1}+c $
B) $ \frac{{{\sec }^{p}}x}{p}+c $
C) $ \frac{{{\tan }^{p+1}}x}{p+1}+c $
D) $ \frac{{{\tan }^{p}}x}{p}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ \sec x=t\Rightarrow \sec x\tan x,dx=dt, $ therefore $ \int_{{}}^{{}}{{{\sec }^{p}}x\tan x,dx}=\int_{{}}^{{}}{{t^{p-1}}dt=\frac{t^{p}}{p}+c}=\frac{{{\sec }^{p}}x}{p}+c. $
BETA
