Integral Calculus Question 437
Question: . $ \int_{{}}^{{}}{{{\tan }^{4}}x\ dx=} $
Options:
A) $ {{\tan }^{3}}x-\tan x+x+c $
B) $ \frac{1}{3}{{\tan }^{3}}x-\tan x+x+c $
C) $ \frac{1}{3}{{\tan }^{3}}x+\tan x+x+c $
D) $ \frac{1}{3}{{\tan }^{3}}x+\tan x+2x+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{{{\tan }^{4}}x,dx}=\int_{{}}^{{}}{{{\tan }^{2}}x({{\sec }^{2}}x-1),dx} $ $ =\int_{{}}^{{}}{{{\tan }^{2}}x{{\sec }^{2}}x,dx}-\int_{{}}^{{}}{{{\tan }^{2}}x,dx}=\frac{{{\tan }^{3}}x}{3}-\tan x+x+c $ .
BETA
