Question

How do you integrate `sec(5/2x) tan(5/2x) dx`?

Answer 1

Personnaly i don't use the `sec(x)` function

`u = 5/2x`
`du = 5/2`

`int1/cos(5/2x)*sin(5/2x)/cos(5/2x)dx = 2/5int1/cos(u)*sin(u)/cos(u)du`

`2/5intsin(u)/cos^2(u)du =-2/5int- sin(u)*cos^-2(u)du`

We have : `u'*u^n`

Remember the formula `intu'u^n = 1/(n+1)*u^(n+1)`

`-2/5int1/(-2+1)*cos^(-2+1)(u) = -2/5int-1/cos(u)du`

`=>2/5[1/cos(5/2x)]+C`