Question

integrate: cot^-1(x) dx

Answer 1

u sure it is integrable?

Answer 2

yep; just a pain in the butt to do i believe :)

Answer 3

u sure it is integrable?

Answer 4

guess so

Answer 5

int by parts using the cot-1 as the deriving part; and dx is the integrating part

Answer 6

.... man these are tricky lol

Answer 7

\begin{array}c
&&dx&\\
+&cot^{-1}(x)&x\\
-&Dx(cot^{-1}(x))&x^2/2
\end{array}

Answer 8

\[cot^{-1}(x)=y\]
\[x=cot(y)\]
\[1=-csc^2(y)dy/dx\]
\[-\frac{1}{csc^2(y)}=y'\]

Answer 9

csc(y) = sqrt(x^2+1)

csc^2(y) = x^2 + 1

Answer 10

is the question: \[\int\limits_{}^{}\cot^{-1} xdx\]

Answer 11

\[Dx(cot^{-1}(x))=-\frac{1}{x^2+1}\]

Answer 12

yes

Answer 13

cot^-1(x) = sinx/cosx?

Answer 14

not quite, the inverse cotangent function doesnt equation the tangent function :)

Answer 15

doesnt *equal* the .....

Answer 16

\[[uv]'=u\ dv+v\ du\]
\[uv =\int u\ dv+\int v\ du\]
\[ \int u\ dv\ =\ uv\ -\int v\ du\]
\[\int cot^{-1}(x)\ dx\]
\[u=cot^{-1}(x);\ dv = dx\]
\[du=-\frac{1}{x^2+1}dx; \ v = x\]
\[ \int cot^{-1}(x)\ dx\ =\ cot^{-1}(x)\ x\ +\int \frac{x}{x^2+1}dx\]
\[=x\ cot^{-1}(x) +\frac{1}{2}\ ln(x^2+1)+C\]
right?

Answer 17

yeah that's good!