Integral Calculus Question 164
Question: What is $ \int{{{\sec }^{n}}x\tan xdx} $ equal to?
Options:
A) $ \frac{{{\sec }^{n}}x}{n}+c $
B) $ \frac{{{\sec }^{n-1}}x}{n-1}+c $
C) $ \frac{{{\tan }^{n}}x}{n}+c $
D) $ \frac{{{\tan }^{n-1}}x}{n-1}+c $ Where ?c? is a constant of integration.
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ I=\int{{{\sec }^{n}}x\tan xdx.} $ Put, $ \sec x=t\Rightarrow \sec x\tan xdx=dt $
$ \therefore I=\int{t^{n}.\frac{dt}{t}} $ $ =\int{{t^{n-1}}dt=\frac{t^{n}}{n}+c=\frac{{{\sec }^{n}}x}{n}+c} $ Where ?c? is a constant of integration.
BETA
