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

1 Answer
Jan 13, 2018

The answer is =1/(2(1-sin^2x))+C

Explanation:

Reminder :

intx^ndx=x^(n+1)/(n+1)+C(n!=-1)

Apply tanx=sinx/cosx

Therefore,

int(tanxdx)/cos^2x=int(sinxdx)/(cos^3x)

Perform the substitution

u=cosx, =>, du=-sinxdx

Therefore,

int(tanxdx)/cos^2x=-int(du)/u^3=-intu^-3du

=-(-u^-2/2)

=1/(2cos^2x)+C

=1/(2(1-sin^2x))+C