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) \[\frac{1}{3}{{\tan }^{3}}x+\tan x+2x+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{{{\tan }^{4}}x\,dx}=\int_{{}}^{{}}{{{\tan }^{2}}x({{\sec }^{2}}x-1)\,dx}\] \[=\int_{{}}^{{}}{{{\tan }^{2}}x{{\sec }^{2}}x\,dx}-\int_{{}}^{{}}{{{\tan }^{2}}x\,dx}=\frac{{{\tan }^{3}}x}{3}-\tan x+x+c\].You need to login to perform this action.
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