Question

Evaluate the integral? : `int \ sec^3x \ dx`

Answer 1

`int \ sec^3x \ dx = 1/2secxtanx - 1/2ln|secx+tanx| + C`

Explanation:

We seek:

`I = int \ sec^3x \ dx`

We can write this as:

`I = int \ sec x \ sec^2x \ dx`

And now we can apply Integration By Parts:

Let `{ (u,=secx, => (du)/dx,=secxtanx), ((dv)/dx,=sec^2x, => v,=tanx ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (secx)(sec^2x) \ dx = (secx)(tanx) - int \ (tanx)(secxtanx) \ dx`

`:. I = secxtanx - int \ secxtan^2x \ dx`
`\ \ \ \ \ \ \ = secxtanx - int \ secx(sec^2x-1) \ dx`
`\ \ \ \ \ \ \ = secxtanx - int \ sec^3x- int \ secx \ dx`
`\ \ \ \ \ \ \ = secxtanx - I - ln|secx+tanx| + A`
`:. 2I = secxtanx - ln|secx+tanx| + A`

Hence:

`:. I = 1/2secxtanx - 1/2ln|secx+tanx| + C`