Question

u substitution

Answer 1

\[\int\limits_{}^{}\cot^23xdx\]

Answer 2

@zepdrix

Answer 3

Do you remember your Half-Angle Identity for Cosine?

Answer 4

can i do \[\int\limits_{}^{}\frac{ \cos^23x }{ \sin^23x }\]

Answer 5

and take u=sin 3x

Answer 6

No, you've introduced a 1/sin^2(3x) without balancing it. no good :c

Answer 7

Here is your Half-Angle Identity\[\huge \cos^2x=\frac{1}{2}(1+\cos2x)\]

Answer 8

It gets rid of the SQUARE on the cosine, which is what we really really wanted.

Answer 9

You just have to be careful in your problem, because your angle isn't x, it's 3x.\[\huge \cos^2\color{salmon}{3x}=\frac{1}{2}(1+\cos2\cdot \color{salmon}{3x})\]

Answer 10

so should sub in to cos^23x and then solve

Answer 11

Yes, make the substitution I just described above, and your integral will become a bit easier. We should have this,\[\huge \int\limits \cos^2 3x \; dx \quad = \quad \frac{1}{2}\int\limits 1+\cos(2\cdot3x)\; dx\]

Answer 12

hmmm but the question asks for cot^23x right why did u give me cos^23x is that right ot i am wrong

Answer 13

Oh..........
My failure to read correctly :C

Answer 14

Yes you can write it the way you described at the start, with the sine on the bottom. That is the best way to approach it I suppose.

Sorry about that :C grrr

Answer 15

i used that but i am here\[\int\limits_{}^{} \cos3x u^-2 du\]

Answer 16

that one is u^-2

Answer 17

Oh Hmm you're right, we're stuck with our extra cosine aren't we?
We might have to take another approach, that stinks.

Sorry my OpenStudy keeps freezing >:C

Answer 18

its ok

Answer 19

\[\large \cot^2x+1=\csc^2x\]Solving for cot^2x gives us,\[\large \cot^2 x=\csc^2x-1\]
\[\large \int\limits\limits \cot^2 3x dx \quad = \quad \int\limits \csc^2 3x - 1 \;dx\]

Answer 20

Did the directions say to use a U-sub? Because this doesn't seem like that type of problem, hmm.