Question

Integration BP.....
(I try not to just use the formulas for the derivative of inverse trig functions)

(Integral x^2arccot(x) dx )

Answer 1

\[\large \int\limits_{ }^{ }x^2\cot^{-1}(x)~dx\]So, firstly,\[\frac{dy}{dx}(y=\cot^{-x})\]\[\frac{dy}{dx}(\cot y=x)\]\[deriving,~~~~~\frac{dy}{dx}\times(-\csc^2y)=1\]\[\frac{dy}{dx}\times(-\cot^2y-1)=1\]\[\frac{dy}{dx}\times(-x^2-1)=1\]so the derivative of arccot of x, is\[-\frac{1}{x^2+1}\] (just same as arcTAN, but negative)

Now, the integral should come along.
\[\large \int\limits_{ }^{ }x^2\cot^{-1}x~dx=\frac{1}{3}x^3\cot^{-1}x~+~\frac{1}{3}\int\limits_{ }^{ }\frac{x^3}{x^2+1}dx\]
applying u substitution to the second integral,\[u=x^2~~~~\frac{1}{2}du=x~dx\](obviously, counting another "u" in the x^3)
\[\large \int\limits\limits_{ }^{ }x^2\cot^{-1}x~dx=\frac{1}{3}x^3\cot^{-1}x~+~\frac{1}{6}\int\limits\limits_{ }^{ }\frac{u}{u+1}du\]
then substitution of w.
\[w=u+1,~~~~~u=w-1,~~~~~dw=du.\]\[\large \int\limits\limits_{ }^{ }x^2\cot^{-1}x~dx=\frac{1}{3}x^3\cot^{-1}x~+~\frac{1}{6}\int\limits\limits_{ }^{ }\frac{w-1}{w}dw\]\[\large \int\limits_{ }^{ }x^2\cot^{-1}x~dx=\frac{1}{3}x^3\cot^{-1}x+\frac{1}{6}w-\frac{1}{6}\ln(w)+C\]
now substitution back,
\[\large \int\limits_{ }^{ }x^2\cot^{-1}x~dx=\frac{1}{3}x^3\cot^{-1}x+\frac{1}{6}(u-1)-\frac{1}{6}\ln(u-1)+C\]\[\large \int\limits_{ }^{ }x^2\cot^{-1}x~dx=\frac{1}{3}x^3\cot^{-1}x+\frac{1}{6}(x^2-1)-\frac{1}{6}\ln(x^2-1)+C\]a small rearrangement,
\[\large \int\limits_{ }^{ }x^2\cot^{-1}x~dx=\frac{1}{3}x^3\cot^{-1}x+\frac{1}{6}x^2-\frac{1}{6}\ln(x^2-1)-\frac{1}{6}+C\]\[\large \int\limits_{ }^{ }x^2\cot^{-1}x~dx=\frac{1}{3}x^3\cot^{-1}x+\frac{1}{6}x^2-\frac{1}{6}\ln(x^2-1)+C\]

Answer 2

@jim_thompson5910

Answer 3

you let w = u+1

but then you made replacements based on the idea that w = u-1. So there's a contradiction

Answer 4

so it should be just the ln(x^2+1), and the +1/6... which is just a vertical shift so I "took it out" ?

Answer 5

yeah and I think the +1/6 kinda gets absorbed by the +C

Answer 6

yes, .. I meant to say what you said, "gets absorbed by C" I'll borrow this one from you for my future reference...

ty so much !!

Answer 7

you're welcome