Question

What is `int "arccot"x dx`?

Answer 1

The answer is `=x arc cotx+1/2ln(x^2+1)+C`

Explanation:

Perform an integration by parts

`intuv'=uv-intu'v`

Here,

`u=arc cotx`, `=>`, `u'=-1/(x^2+1)`

`v'=1`, `=>`, `v=x`

The integral is

`I=x arc cotx+int(xdx)/(x^2+1)`

`=x arc cotx+1/2int(2xdx)/(x^2+1)`

`=x arc cotx+1/2ln(x^2+1)+C`

Answer 2

`x cot^-1x + 1/2ln(x^2 + 1) + C`

Explanation:

We have,

`int cot^-1x dx`

Integrate by Parts. [`intuvdx = uintvdx - int(d/dx(u)intvdx)`]

`= cot^-1x int dx - int (d/dx(cot^-1x)intdx)dx`

`= xcot^-1x - int(-1/(x^2 + 1) * x)dx`

`= xcot^-1x - intx/(-x^2 - 1)dx`.........................(i)

Now, Let's Solve `int-x/(x^2 + 1) dx`.

`int x/(-x^2 - 1) dx`

Substitute `u = -x^2 - 1`.

That Means, `du = -2x dx rArr dx = -1/(2x) du`

So,

`int x/(-x^2 - 1) dx`

`= int x/u dx`

`= int cancelx/u * (-1/(2cancelx))du`

`= -1/2 int 1/u du`

`= -1/2 ln|u| + C`

`= -1/2 ln|-x^2 - 1| + C`....................(ii)

Now, From (i),

`int cot^-1x dx`

`= xcot^-1x + 1/2ln|-x^2 - 1| + C`

`= xcot^-1x + 1/2ln(x^2 + 1) + C`

And, That's settled.

Hope this helps.

Answer 3

`intcot^-1(x)"d"x=xcot^-1(x)-lnsin(cot^-1(x))+"c"`

Explanation:

We use the integral of inverse functions theorem:

`intf^-1(x)"d"x=xf^-1(x)+F(f^-1(x))+"c"`

where `F(x)=intf(x)"d"x`

If `f^-1(x)="arccot"x=cot^-1(x)` then `f(x)=cotx` and `F(x)=lnabssinx` (Try to prove this)

So,

`intcot^-1(x)"d"x=xcot^-1(x)-lnsin(cot^-1(x))+"c"`

Note that this is equivalent to the other answer because

`sin(cot^-1(x))=1/csc(cot^-1(x))=1/sqrt(cot^2(cot^-1(x))+1)=1/sqrt(x^2+1)`

so

`-lnsin(cot^-1(x))=-ln(1/sqrt(x^2+1))=-ln((x^2+1)^(-1/2) )=1/2ln(x^2+1)`