How to solve properly #int tan(x)/cos^2x#?
I have a doubt on how to solve this integral.
The first and the correct way to solve is:
#int (tan x) /cos^2x dx#
Let's rewrite
#int sin x / cos^3x dx#
With the #u# substitution:
#u = cos x# , so #du = -sin x#
#-int1/u^3du# = #1/(2u^2) + C#
Substitute the original value and the result is:
#1/(2cos^2x)+C#
The second way with the mistake is:
#int (tan x) /cos^2x dx#
#int (tan x) * 1/cos^2x dx#
#u = tan(x)# , so #du = 1/cos^2x#
Which means:
#int u \ du# = #u^2/2+C#
#tan^2x/2 + C#
I have a doubt on how to solve this integral.
The first and the correct way to solve is:
Let's rewrite
With the
Substitute the original value and the result is:
The second way with the mistake is:
Which means:
1 Answer
The answer is
Explanation:
Reminder :
Apply
Therefore,
Perform the substitution
Therefore,