Question

How do you find `int sec^2x/(1-sin^2x) dx`?

Answer 1

Shown below

Explanation:

`sin^2 x + cos^2 x = 1 => cos^2x = 1 - sin^2 x`

`=> int sec^2x / ( cos^2 x ) dx`

`=> int sec^2 x * 1/cos^2 x dx`

`=> int sec^4 x dx`

`=> int sec^2 x * sec^2 x dx`

`=> int ( 1 + tan^2x) * sec^2 x dx`

`u = tanx`

`du = sec^2 x dx`

`=> int 1 + u ^2 du`

`=> u + 1/3 u^3 + c`

`= tanx + 1/3 tan^3 x + c`

Answer 2

`tan(x) + frac(1)(3) tan^(3)(x) + C`

Explanation:

We have: `int` `frac(sec^(2)(x))(1 - sin^(2)(x))` `dx`

`= int` `frac(sec^(2)(x))(cos^(2)(x))` `dx`

`= int` `frac(sec^(2)(x))(frac(1)(sec^(2)(x)))` `dx`

`= int` `sec^(2)(x) cdot sec^(2)(x)` `dx`

Then, the Pythagorean identity is `cos^(2)(x) + sin^(2)(x) = 1`.

We can divide through by `cos^(2)(x)` it to get:

`Rightarrow 1 + tan^(2)(x) = sec^(2)(x)`

Let's apply this rearranged identity to our integral:

`= int` `(1 + tan^(2)(x)) cdot sec^(2)(x)` `dx`

Now, let's use `u`-substitution, where `u = tan(x) Rightarrow du = sec^(2)(x)` `dx`:

`= int` `(1 + u^(2))` `du`

`= int` `1` `du + int` `u^(2)` `du`

`= u + frac(1)(3) u^(3) + C`

Finally, we can substitute `tan(x)` in place of `u`:

`= tan(x) + frac(1)(3) tan^(3)(x) + C`