Question

Evaluate the integral by using multiple substitutions

∫ [(2r-2)cos√(3(2r-2)^2+9)]/[√(3(2r-2)^2+9)] [dr]

Answer 1

are you familiar with trigonometric substitution?

Answer 2

Nope, I'm not good at it. Is this problem similar to the one I asked before?

Answer 3

i believe so.

Answer 4

we know that tan(x)^2+1=sec(x)^2

Answer 5

yup

Answer 6

and if \[x=a*tan(\theta) \rightarrow \sqrt{x^2-a^2}=a*sec(\theta)\]
this means \[\sqrt{3(2r-2)^2+9}=\sqrt{3}*\sqrt{(2r-2)^2+3}\]

Answer 7

so we say \[(2r-2)=\sqrt{3}*tan(\theta)\]

Answer 8

which means we have
\[\sqrt{3(2r−2)^2+9}=\sqrt{3}*\sqrt{(2r−2)^2+3}=\sqrt{3}*\sqrt{3}sec(\theta)=3sec(\theta)\]

Answer 9

does this make sense so far?

Answer 10

yeah its good. I kinda go lost when you said it was (2r-2)=Sqroot(3)*tan bc i didn't understand where the +9 and +3 went but now i'm good

Answer 11

*got

Answer 12

I get it :)

Answer 13

ok :)

Answer 14

so now we can substitute our tan and sec functions into the problem.

Answer 15

\[\frac{(2r-2)cos(\sqrt{(3(2r-2)^2+9)}}{\sqrt{(3(2r-2)^2+9)}}\]
becomes
\[\frac{\sqrt{3}tan(\theta)cos((3sec(\theta))}{3sec(\theta)}\]

Answer 16

kk TY I'll post the answer once i'm done with the problem ...^_^

Answer 17

Ohh XO TY I need to practice these kinds of problems more :/

Answer 18

we aren't done with this substitution part yet.
there is a vital step we still have left.

Answer 19

you have to solve for dr in terms of d(theta)

Answer 20

\[(2r−2)=\sqrt{3}tan(θ)\]
\[r=\frac{\sqrt{3}tan(θ)+2}{2}\]
\[dr=\frac{\sqrt{3}}{2}sec^2(\theta)\]

Answer 21

whoops i mean
\[dr=\frac{\sqrt{3}}{2}sec^2(\theta)d\theta\]

Answer 22

oh ok lol the theta part went over my head XP

Answer 23

so
\[\int\frac{(2r-2)cos(\sqrt{(3(2r-2)^2+9)}}{\sqrt{(3(2r-2)^2+9}} dr =\int\frac{\sqrt{3}tan(\theta)cos(3sec(\theta))}{3sec(\theta)} *\frac{\sqrt{3}}{2}sec^2(θ)dθ\]
\[=\int\frac{sec(\theta)tan(\theta)cos(3sec(\theta))}{2} dθ\]

Answer 24

well our orinial integral uses dr
we need dθ because our funtion is writen in terms of θ

Answer 25

so if we solve for r in terms of θ and take the derivative we found dr in terms of θ and dθ which is what we needed

Answer 26

make sense?

Answer 27

yes I understand. I'll do it over again bc what I had before is all crossed out XP

Answer 28

Thank you! Didn't think this problem would be so long T.T

Answer 29

we have a second substitution still...

Answer 30

we can't integrate cos(3sec(θ))

Answer 31

we can integrate cos(u) though.

Answer 32

XO my teacher is nuts. LOL but yes I understand bc it can't stay with theta

Answer 33

so
\[u=3sec(\theta)\]
\[du=3sec(\theta)tan(\theta)\]

Answer 34

i mean
\[du=3sec(\theta)tan(\theta)d\theta\]

Answer 35

so
\[\int \frac {sec(θ)tan(θ)cos(3sec(θ))}{2}dθ=\frac{1}{3}\int \frac {cos(3sec(θ))}{2}*3sec(θ)tan(θ)dθ\]
\[=\frac{1}{3}\int \frac{cos(u)}{2}du=\frac16sin(u)\]

Answer 36

\[u=3sec(\theta)\]
\[\theta=tan^{-1}(\frac{\sqrt{3}(2r-2)}{3})\]

Answer 37

\[\frac16sin(u)=\frac16sin(3sec(θ))=\frac16sin(3sec(tan^{-1}(\frac{\sqrt{3}(2r-2)}{3})))\]

Answer 38

and that is the final answer :) do you get where each part comes from?

Answer 39

yes I do, but I'm still writing it down and writing side notes to help me know where certain things came from where. Wow its really long thank you so much for your time and patience! I appreciate it =)

Answer 40

of course :)

Answer 41

You did an amazing job! You deserve lots of medals and a good rest =)

Answer 42

thanks :)