What is the antiderivative of #y=arc cotx#?

Question

1277 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-18T20:59:42

#int"arccot"(x)dx=x"arccot"(x)+1/2ln(x^2+1)+C#.

Let #I=intydx=int"arccot"(x)dx=int("arccot"(x))(1)dx#

Now, we use the Rule of Integration by Parts, which states,

#intuvdx=uintvdx-int((du)/dx*intvdx)dx#

We take #u="arccot"(x)rArr (du)/dx=-1/(x^2+1)#, and,

#v=1 rArr intvdx=x#.

Hence, #I=x"arccot"(x)+int x/(x^2+1)dx#

#=x"arccot"(x)+1/2int(2x)/(x^2+1)dx#

#=x"arccot"(x)+1/2int(d(x^2+1))/(x^2+1)#

#:. I=x"arccot"(x)+1/2ln(x^2+1)+C, .......[as, intdt/t=ln|t|]#.