Question

How do you integrate `int tan^3((pix)/2)sec^2((pix)/2)`?

Answer 1

`inttan^3((pix)/2)sec^2((pix)/2)dx=tan^4((pix)/2)/(2pi)+C`

Explanation:

`I=inttan^3((pix)/2)sec^2((pix)/2)dx`

We can first eliminate some ugliness by letting `u=(pix)/2`. Differentiating this gives `du=pi/2dx`, so rearrange the integral as follows:

`I=2/piinttan^3((pix)/2)sec^2((pix)/2)(pi/2dx)`

`I=2/piinttan^3(u)sec^2(u)du`

Notice that `sec^2` is the derivative of `tan`, so let `v=tan(u)` so `dv=sec^2(u)du`:

`I=2/piintv^3dv`

`I=2/piv^4/4`

`I=(v^4)/(2pi)`

`I=tan^4(u)/(2pi)`

`I=tan^4((pix)/2)/(2pi)+C`