Question

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`

Answer 1

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`