Question

Calc 2 problem someone mind lending a hand?

Answer 1

\[\int\limits (\cos x)[(\ln(\sin(x)]/(\sin x)\]

Answer 2

what did you think about so far
looks a good substitution no?

Answer 3

is this what you have \[\int \frac{\cos x \ln(\sin x) }{\sin x }dx\]

Answer 4

\(u=\sin x ~~~~~~\Longrightarrow ~~~~ du=\cos x dx\)

Answer 5

so now we get this integral after sub
\[\int \frac{\ln u}{u}~~~du\]

Answer 6

yet another sub for this new integral \(t=\ln u ~~~\Longrightarrow ~~~dt=\frac{1}{u}du\)

Answer 7

i leave you with that.... finish it up

Answer 8

what does the new integrand look like?

Answer 9

answer: \[\ln^2(\sqrt{\sin x })\]
see how it was found

Answer 10

+C of course i forgot that

Answer 11

actually error lol
answer is :
\[\frac{\ln^2(\sin x)}{2}+c\]

Answer 12

nice! thanks. i just needed to know which technique would be acceptable in solving this, forgot how to derive natural logs but i see now. thanks a ton!

Answer 13

U sub was nice with this :)

Answer 14

welcome

Answer 15

yup. issue i find is determining which technique to use. i guess you just try to use them all and see which one seems to work the best or may be the only one that works?

Answer 16

try to solve this \[\int\sqrt{\frac{x}{1-x^3}}~~dx \]

Answer 17

well it is about seeing what can work. i looked at sin
and saw that cos at the top is its derivative so i knew substitution would work great
once you do this more and more you develop the sense of knowing what works and what not

Answer 18

Training is the best way to do this :) more and more problems

Answer 19

uhhh i'd imagine we'd rewrite 1-x^3 using trig sub?

Answer 20

i would go the whole hog and let \(u=\log(\sin(x))\) making \(du=\frac{\cos(x)}{\sin(x)}dx\)

Answer 21

but if you differentiate that you get -3x^2 , which is something that you don't have

Answer 22

yeah thats why i was mentioning remembering derivative of nautral logs haha @satellite73

Answer 23

yeah true :) i just realized that lol

Answer 24

@satellite73 wild sense of eyeballing

Answer 25

so 1-x^3 using trig sub or was that a no? was thinking we could say 1-x^2 * x then rewrite 1-x^2 using x = sin theta perhaps?