Question

Evaluate `I=\int\sec^3(x)dx`?

The problem suggests using integration by parts.
After splitting the integral into `\int\sec^2(x)\sec(x)dx`, I tried `u=\sec(x)` and `dv=\sec^2(x)dx`...

But I am stuck on the resulting `uv-\intvdu` form (here is the equation I got).

Answer 1

`int sec^3x dx = (secxtanx)/2 + 1/2ln abs (secx+tanx)+C`

Explanation:

Note that:

`d/dx tanx = sec^2x`

`d/dx secx = secx tanx`

so:

`int sec^3x dx = int secx * sec^2x dx = int secx d(tanx)`

so, integrating by parts:

`int sec^3x dx = secxtanx - int tanx d(secx)`

`int sec^3x dx = secxtanx - int secx tan^2xdx`

Use now the trigonometric identity:

`tan^2x = sec^2x -1`

to have:

`int sec^3x dx = secxtanx - int secx (sec^2x-1)dx`

and using the linearity of the integral:

`int sec^3x dx = secxtanx - int sec^3xdx + int secxdx`

The integral to solve now appears on both sides of the equation, then:

`2int sec^3x dx = secxtanx + int secxdx`

`int sec^3x dx = (secxtanx)/2 + 1/2 int secxdx`

You can see here how to solve the resulting integral to have:

`int sec^3x dx = (secxtanx)/2 + 1/2ln abs (secx+tanx)+C`

Answer 2

`I=1/2tanxsecx+1/2ln|tanx+secx|+c`

Explanation:

Here,

`I=intsec^3xdx`

`=intsecxsec^2xdx`

`=intsqrt(1+tan^2x)sec^2xdx`

Let , `tanx=u=>sec^2xdx=du`

`I=intsqrt(1+u^2)du`

`=u/2sqrt(1+u^2)+1/2ln|u+sqrt(u^2+1)|+c`,where, `u=tanx`

`=1/2tanxsecx+1/2ln|tanx+sqrt(1+tan^2x)|+c`

`I=1/2tanxsecx+1/2ln|tanx+secx|+c`