Question

What is `int tan^3(x) sec^4(x/2) dx`?

Answer 1

`8/(1-tan^2(x/2))^2-24/(1-tan^2(x/2))-8ln(abs(1-tan^2(x/2)))+C`

Explanation:

The first step here should be to eliminate the half-angle--even though it will crate a double angle in the tangent function, double-angles are easier to work with.

Let `u=x/2`, such that `x=2u` and `du=1/2dx` and `dx=2du`.

Substituting these in, we see that:

`inttan^3(x)sec^4(x/2)dx=2inttan^3(2u)sec^4(u)du`

Note the tangent double-angle formula:

`tan(2u)=(2tan(u))/(1-tan^2(u))`

Therefore:

`2inttan^3(2u)sec^4(u)du=2intsec^4(u)[(2tan(u))/(1-tan^2(u))]^3du`

Now, cube the fraction and take a `sec^2(u)` from the `sec^4(u)` and write it as `1+tan^2(u)`. This gives us an integrand of entirely tangents, except for a `sec^2(u)`, its derivative.

`=2intsec^2(u)((8tan^3(u)(1+tan^2(u)))/(1-tan^2(u))^3)du`

Now, let `v=tan(u)` such that `dv=sec^2(u)du`.

`=16int(v^3(1+v^2))/(1-v^2)^3dv`

Now, let `w=1-v^2` so that `dw=-2vdv`. Rearranging so that `-2v` is present:

`=-8int(-2v(v^2)(1+v^2))/(1-v^2)^3dv`

Here, from `w=1-v^2`, we see that `v^2=1-w` and `1+v^2=2-w`. Hence:

`=-8int((1-w)(2-w))/w^3dw`

Expanding:

`=-8int(2-3w+w^2)/w^3dw`

Splitting and writing with negative exponents, except for on the last one (since it's the natural logarithm integral:

`=-16intw^-3dw+24intw^-2-8int1/wdw`

Integrating using `intw^ndw=w^(n+1)/(n+1)+C` and `int1/wdw=ln(absw)+C`, we see that:

`=-16(w^-2/(-2))+24(w^-1/(-1))-8ln(absw)+C`

Using `w=1-v^2`:

`=8/(1-v^2)^2-24/(1-v^2)-8ln(abs(1-v^2))+C`

Using `v=tan(u)`:

`=8/(1-tan^2(u))^2-24/(1-tan^2(u))-8ln(abs(1-tan^2(u)))+C`

Using `u=x/2`:

`=8/(1-tan^2(x/2))^2-24/(1-tan^2(x/2))-8ln(abs(1-tan^2(x/2)))+C`