Integral Calculus Question 435
Question: $ \int_{{}}^{{}}{e^{x}{{\tan }^{2}}(e^{x})dx=} $
Options:
A) $ \tan (e^{x})-x+c $
B) $ e^{x}(\tan e^{x}-1)+c $
C) $ \sec (e^{x})+c $
D) $ \tan (e^{x})-e^{x}+c $
Show Answer
Answer:
Correct Answer: D
Solution:
Put $ e^{x}=t\Rightarrow e^{x}dx=dt, $ then $ \int_{{}}^{{}}{e^{x}{{\tan }^{2}}(e^{x}),dx}=\int_{{}}^{{}}{{{\tan }^{2}}t,dt}=\int_{{}}^{{}}{({{\sec }^{2}}t-1),dt} $ $ =\tan t-t+c=\tan (e^{x})-e^{x}+c. $
BETA
