Question

How do you find the integral of `f(x)=(sec^2 x)(tan^2 x)` using integration by parts?

Answer 1

Apply integration by parts to generate the same integral, and then solve the equation algebraically.

Explanation:

In integration by parts, we use the formula

`intudv = uv - intvdu`

In this case, we will set `u = tan^2(x)` and `dv = sec^2(x)dx`

Then `du = 2tan(x)sec^2(x)dx` and `v = tan(x)`.

(Note that we chose `u` and `dv` in such a way as to make sure that `du` and `v` were easy to calculate. This does not always work, but is often a good place to start.)

Now, let `I = intudv = intsec^2(x)tan^2(x)dx`

Applying the integration by parts formula, we get
`intsec^2(x)tan^2(x)dx = tan^2(x)tan(x) - inttan(x)(2tan(x)sec^2(x))dx`

But some simplification of the right hand side gives us

`intsec^2(x)tan^2(x)dx = tan^3(x) - 2intsec^2(x)tan^2(x)dx`

Now, we add `2intsec^2(x)tan^2(x)dx` to both sides to obtain

`3intsec^2(x)tan^2(x)dx = tan^3(x)`

Dividing by 3 and adding in the constant gives us our final answer

`intsec^2(x)tan^2(x)dx=1/3 tan^3(x) + C`

---

.
.
.

---

The following is unnecessary for understanding the above solution, and simply addresses a technical detail.

In the work above, it seems as if the constant disappears during the algebra and we need to add it in manually at the end.
To avoid the issue with seemingly canceling the constant, we can note that `intsec^2(x)tan^2(x)dx = g(x) + C` for some `g(x)`
Now, we can write the two instances of `intsec^2(x)tan^2(x)dx` with arbitrary constants, giving us
`g(x) + C_1 = tan^3(x) - 2(g(x) + C_2)`

Solving for g(x) gives
`g(x) = 1/3tan^3(x)- 1/3(C_1 + 2C_2)`

Substituting gives us
`intsec^2(x)tan^2(x)dx = 1/3tan^3(x)- 1/3(C_1 + 2C_2) + C`

But as `- 1/3(C_1 + 2C_2)` is itself an arbitrary constant, and the sum of arbitrary constants is an arbitrary constant, it is equivalent and simpler to have `C` encompass `- 1/3(C_1 + 2C_2)` and write

`intsec^2(x)tan^2(x)dx=1/3 tan^3(x) + C`