Question

int cot^5dx

Answer 1

\[\cot = \frac{ \cos }{ \sin }\]

Answer 2

factor out a sin or cos to use and substitute.

Answer 3

We'll need to use this important identity,\[\large \cot^2x+1=\csc^2x \qquad \rightarrow \qquad \cot^2x=\csc^2x-1\]

\[\large \int\limits \cot^5x\;dx \qquad = \qquad \int\limits \cot^3x(\csc^2x-1)\;dx\]
Ok here is the first step :) Understand what's going on so far?

I don't think that method will work abbot, since you're not able to get rid of all the cosines on top. Your du will only take care of one of them :(

Answer 4

ok thanx

Answer 5

Then distribute the cot^3x to each term in the brackets, and split it into 2 integrals.
\[\large \int\limits\limits \cot^3x(\csc^2x-1)\;dx \qquad = \qquad \int\limits \color{royalblue}{\cot^3x \csc^2xdx}-\int\limits \cot^3x dx\]

This blue one is now ready for a `u sub`.
Let \(\large u=cot^3x\).

Answer 6

Woops, let \(\large u=cot x\). Typo there :P

Answer 7

(int)cot^3x*csc^2x dx - (int)cot^3xdx
(int) (csc^2x-1)*csc^2x cotx dx -(int)cot^3x
(int)csc^4x - csc^2x Cotx dx (int)cot^3x dx
1/5 csc^5x - 1/3 csc^3 + 1/4 csc^4 + c
this the answer right :)

Answer 8

:( thanx any way

Answer 9

Let \(\large \color{royalblue}{u=\cot x}\)
Then \(\large du=-\csc^2x\;dx\qquad\rightarrow\qquad \color{orangered}{-du=\csc^2x\;dx}\)

This is how the first portion will change when you apply a `U sub`.
\[\large \int\limits\limits\limits (\color{royalblue}{\cot x})^3 \color{orangered}{\csc^2xdx} \qquad \rightarrow \qquad \int\limits (\color{royalblue}{u})^3(\color{orangered}{-du})\]

Understand? :o