Question

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

Answer 1

`inttan^2(x)sec^3(x)dx = (2sec^3(x)tan(x) - sec(x)tan(x)-ln|sec(x)+tan(x)|)/8`

Explanation:

`I = inttan^2(x)*sec^3(x)dx`
`I = int(sec^2(x)-1)*sec^3(x)dx`
`I = intsec^5(x)dx - intsec^3(x)dx`

For the second integral, you need integration by parts, say
`u = sec(x)`, so `du = sec(x)tan(x)` and
`dv = sec^2(x)` so `v = tan(x)`

`intsec^3(x)dx = sec(x)tan(x) - intsec(x)tan^2(x)dx`
`intsec^3(x)dx = sec(x)tan(x) - intsec(x)(sec^2(x)-1)dx`
`intsec^3(x)dx = sec(x)tan(x) - int(sec^3(x)-sec(x))dx`
`intsec^3(x)dx = sec(x)tan(x) - intsec^3(x)dx -intsec(x)dx`
`2intsec^3(x)dx = sec(x)tan(x) - intsec(x)dx`

The last integral is a tabled one, so

`intsec^3(x)dx = (sec(x)tan(x)+ln|sec(x)+tan(x)|)/2 + c`

For `intsec^5dx`, using integration by parts
`u = sec^3(x)` so `du = 3tan(x)sec^3(x)`
`dv = sec^2(x)` so `v = tan(x)`

`intsec^5dx = sec^3(x)tan(x) - inttan(x)*3tan(x)sec^3(x)dx`
`intsec^5dx = sec^3(x)tan(x) - 3inttan^2(x)sec^3(x)dx`

But we originally had that

`inttan^2(x)sec^3(x)dx = intsec^5(x)dx - intsec^3(x)dx`

So substituting that we have

`inttan^2(x)sec^3(x)dx = sec^3(x)tan(x) - 3inttan^2(x)sec^3(x)dx - (sec(x)tan(x)+ln|sec(x)+tan(x)|)/2`
`4inttan^2(x)sec^3(x)dx = (2sec^3(x)tan(x) - sec(x)tan(x)-ln|sec(x)+tan(x)|)/2`
`inttan^2(x)sec^3(x)dx = (2sec^3(x)tan(x) - sec(x)tan(x)-ln|sec(x)+tan(x)|)/8`