Question

How do you integrate `tanx / (cosx)^2`?

Answer 1

`inttanx/(cosx)^2dx=1/2tan^2x+C`

Explanation:

Let `u=tanx`, hence `du=sec^2xdx`

Hence `inttanx/(cosx)^2dx`

= `inttanxsec^2xdx`

= `intudu`

= `u^2/2`

= `1/2tan^2x+C`

Answer 2

`sec^2x/2+C" "` or `" "tan^2x/2+C`

Explanation:

`I=inttanx/cos^2xdx`

Since `tanx=sinx/cosx`:

`I=intsinx/cos^3xdx`

We will use the substitution `u=cosx`, which implies that `du=-sinxdx`:

`I=-int(-sinx)/cos^3xdx`

`I=-intu^-3du`

Using the typical power rule for integration:

`I=-(u^-2/(-2))+C`

`I=1/(2u^2)+C`

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

`I=sec^2x/2+C`

This is equivalent to the answer found by Shwetank Mauria, as `sec^2x` and `tan^2x` are separated by a constant: `sec^2x=tan^2x+1`.

`I=(tan^2x+1)/2+C`

`I=tan^2x/2+1/2+C`

The `1/2` is absorbed into `C`;

`I=tan^2x/2+C`

Both answers are valid.