What is int "arccot"x dx?

3 Answers
May 23, 2018

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

May 23, 2018

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.

May 23, 2018

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)