Question

What's the integral of `int tanx / (cosx)^2 dx`?

Answer 1

`int tanx / (cosx)^2 dx = 1/2 sec^2x +C` (Or, equivalently `1/2tan^2x +C`, depending on method used.)

Explanation:

Method 1

`tanx/cos^2x = sinx/cosx 1/cos^2x = sinx (cosx)^-3`

Integrate by substitution with `u=cosx`.

Method 2

`tanx/cos^2x = tanx seec^2x = (secx)(secxtanx)`

Integrate by substitution with `u=secx`.

Method 3

`tanx/cos^2x = tanx seec^2x`

Integrate by substitution with `u=tanx`.

This method gets the antiderivative in the form `1/2tan^2x + C`

Answer 2

I found: `1/(2cos^2(x))+c`

Explanation:

I would write it as:
`int(tan(x))/cos^2(x)dx=intsin(x)/cos(x)1/cos^2(x)dx=intsin(x)/cos^3(x)dx=`
I can use the fact that:
`d[cos(x)]=-sin(x)dx` to write:
`int-(d[cos(x)])/cos^3(x)=int-cos^-3(x)(d[cos(x)])=` where `cos(x)` is our variable of integration; giving:
`=-cos^-2/-2+c=1/(2cos^2(x))+c`