A) \[y=\tan x+1+c{{e}^{-\,\tan x}}\]
B) \[y=\tan x-1+c{{e}^{-\,\tan x}}\]
C) \[y=\tan x-1+c{{e}^{\tan x}}\]
D) None of these
Correct Answer: B
Solution :
\[y\cdot {{e}^{\int{{{\sec }^{2}}x\,dx}}}=\int{{{e}^{\tan \,x}}}\cdot \tan x\cdot {{\sec }^{2}}x\,\,dx+c\] \[y\cdot {{e}^{\tan x}}={{e}^{\tan x}}(\tan x-1)+c\] \[y=(\tan x-1)+c\cdot {{e}^{-\tan x}}\]
You need to login to perform this action.
You will be redirected in
3 sec
