Question

How do you find the integral of `tan^2(x) * sec^3(x) dx`?

Answer 1

See the explanation section, below.

Explanation:

Rewrite the integrand using `tan^2x = sec^2x-1`.

Let's give the integral we want the name `I`

`I = int tan^2xsec^3x dx = int (sec^5x-sec^3x)dx`

Next we'll integrate `sec^5x` by parts.

`int sec^5x dx = int sec^3 x sec^2x dx`

Let `u = sec^3 x` and `dv = sec^2x dx`.

Then `du = 3tanx sec^3x dx` and `v = tanx`

We get

`int sec^5 x dx = sec^3x tanx - 3int tan^2x sec^3x dx`

Again, use `tan^2x = sec^2 x-1` to get

`int sec^5 x dx = sec^3x tanx - 3int (sec^2 x-1) sec^3x dx`

`int sec^5 x dx = sec^3x tanx - 3int sec^5 dx + 3int sec^3x dx`

Which gets us

`4int sec^5 x dx = sec^3x tanx + 3int sec^3x dx`

and

`int sec^5 x dx = 1/4sec^3x tanx + 3/4 int sec^3x dx`

Recalling that the other integral we need is `int sec^3x dx`, let's simplify our lives by writing:

`I = intsec^5xdx-intsec^3xdx`

`= underbrace(1/4sec^3x tanx + 3/4 int sec^3x dx)_(intsec^5 x dx)-intsec^3xdx`

`= 1/4sec^3x tanx -1/4 int sec^3x dx`

So now find `int sec^3x dx = int secx sec^2x dx` using the same general approach and integration by parts. You should get

`int sec^3x dx = 1/2secxtanx - 1/2int secx dx`

So at this point we have

`I = 1/4sec^3x tanx -1/4underbrace( [1/2secxtanx - 1/2int secx dx])_(intsec^3 x dx)`

`= 1/4sec^3x tanx -1/8secxtanx + 1/8 int secx dx`

Now `int secx dx` can be evaluated in a couple of ways, the usual trick is to multiply by `(secx+tanx)/(secx+tanx)` to get `int 1/u du = ln absu = ln abs(secx+tanx)`

So finally we finish with

`I = 1/4sec^3x tanx - 1/8secxtanx + 1/8 ln abs(sec x + tan x ) +C`

Alternative form for `int secx dx`

`intsecx dx` can also be found by substitution and partial fractions to get `1/2ln abs((sinx+1)/(sinx-1))+C`
(Yes, the is equivalent to `ln abs(secx+tanx)+C`)