What is integral of?

#intarctan(cotx)dx#=?

1 Answer
Apr 14, 2018

#intarctan(cot(x))dx=xarctan(cot(x))+x^2/2+C#

Explanation:

We can use integration by parts.

It states that:

#intudv=uv-intvdu#

We let: #u=arctan(cot(x))# and #dv=1#

#=>du=d/dx[arctan(cot(x))]#

#=>du=1/(1+cot^2(x))*d/dx[cot(x)]#

#=>du=1/(1+cot^2(x))*-csc^2(x)#

#=>du=-csc^2(x)/(1+cot^2(x))#

#v=int1dx#

#v=x#

We now have:

#arctan(cot(x))*x-int-csc^2(x)/(1+cot^2(x))*xdx#

#=>arctan(cot(x))*x-int(x*csc^2(x))/(-1-cot^2(x))dx#

Remember that:

#1+cot^2(x)=csc^2(x)# Manimpulate this to get:

#-1-cot^2(x)=-csc^2(x)#

#=>arctan(cot(x))*x-int(x*cancel(csc^2(x)))/(-cancel(csc^2(x)))dx#

#=>arctan(cot(x))*x-int-xdx#

#=>arctan(cot(x))*x+intxdx#

Remember that:

#intx^n=(x^(n+1))/(n+1)#

#=>arctan(cot(x))*x+x^2/2# Do you #C# why this is incomplete?

#=>xarctan(cot(x))+x^2/2+C# That is the answer!