What is integral of?

intarctan(cotx)dxarctan(cotx)dx=?

1 Answer
Apr 14, 2018

intarctan(cot(x))dx=xarctan(cot(x))+x^2/2+Carctan(cot(x))dx=xarctan(cot(x))+x22+C

Explanation:

We can use integration by parts.

It states that:

intudv=uv-intvduudv=uvvdu

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

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

=>du=1/(1+cot^2(x))*d/dx[cot(x)]du=11+cot2(x)ddx[cot(x)]

=>du=1/(1+cot^2(x))*-csc^2(x)du=11+cot2(x)csc2(x)

=>du=-csc^2(x)/(1+cot^2(x))du=csc2(x)1+cot2(x)

v=int1dxv=1dx

v=xv=x

We now have:

arctan(cot(x))*x-int-csc^2(x)/(1+cot^2(x))*xdxarctan(cot(x))xcsc2(x)1+cot2(x)xdx

=>arctan(cot(x))*x-int(x*csc^2(x))/(-1-cot^2(x))dxarctan(cot(x))xxcsc2(x)1cot2(x)dx

Remember that:

1+cot^2(x)=csc^2(x)1+cot2(x)=csc2(x) Manimpulate this to get:

-1-cot^2(x)=-csc^2(x)1cot2(x)=csc2(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!