evaluate

[Ryan Rigdon]

here is a problem that requires us to simplify and type an exact answer using radicals as needed. i am leaning towards 8*ln(3) for my final answer. what do you think?

 

[mmm4444bot]

type an exact answer using radicals as needed ? I'm not sure what "as needed" means, here.

8 * ln(3) is the exact answer.

8.788898 is only an approximation (to six decimal places).

 

[Ryan Rigdon]

just means if we needed to use radicals put them in.

 

[mmm4444bot]

means if we needed to use radicals put them in

You simply repeated the information. Doing that explains nothing!

I still do not understand the condition which satisfies "if we needed to use radicals".

How do you know when you're required to use radical notation versus something else?

(I also do not understand why you're using the past tense. Is the requirement no longer valid?)

 

[BigGlenntheHeavy]

\(\displaystyle You \ are \ all \ wet.\)

\(\displaystyle \int sec(x) csc(x) dx \ \ne \ \int sec(x)dx \ + \ \int csc(x)dx\)

\(\displaystyle (5)(3) \ \ne \ 5+3, \ If \ you \ got \ the \ right \ answer, \ you \ fudged \ it.\)

\(\displaystyle Also \ note, \ \int \frac{1}{cos(x)}dx \ \ne \ ln|sin(x)|\)

 

[BigGlenntheHeavy]

\(\displaystyle Correct \ way \ to \ solve \ this \ definite \ integral.\)

\(\displaystyle \int_{\pi/6}^{\pi/3}8sec(x)csc(x)dx \ = \ 8\int_{\pi/6}^{\pi/3} \frac{dx}{sin(x)cos(x)}\)

\(\displaystyle Now, \ let \ u \ = \ \frac{sin(x)}{1+cos(x)} \ = \ tan(x/2), \ yields \ sin(x) \ = \ \frac{2u}{1+u^2},\)

\(\displaystyle cos(x) \ = \ \frac{1-u^2}{1+u^2}, \ and \ dx \ = \ \frac{2du}{1+u^2}.\)

\(\displaystyle Ergo, \ we \ have \ 8\int_{2-\sqrt3}^{\sqrt3/3}\frac{2du/(1+u^2)}{2u(1-u^2)/(1+u^2)^2} \ = \ 8\int_{2-\sqrt3}^{\sqrt3/3}\frac{1+u^2}{u(1-u^2)}du\)

\(\displaystyle = \ 8\int_{2-\sqrt3}^{\sqrt3/3}\bigg[\frac{1}{u}-\frac{1}{u-1}-\frac{1}{u+1}\bigg]du \ = \ ? \ \ The \ correct \ answer \ is \ 8ln(3).\)

\(\displaystyle Can \ you \ take \ it \ from \ here?\)

 

[Ryan Rigdon]

i see where i fudged it. going to try another way and then try it with your helpful hints. thanx so much will post my findings later.

 

[Ryan Rigdon]

where did sin(x)/(1+cos(x)) come from.

 

[BigGlenntheHeavy]

\(\displaystyle From \ my \ photographic \ memory.\)

\(\displaystyle Note, \ you \ can \ let \ u \ equal \ anything \ as \ long \ as \ everything \ else \ jels.\)

 

[Ryan Rigdon]

\(\displaystyle Correct \ way \ to \ solve \ this \ definite \ integral.\)

\(\displaystyle \int_{\pi/6}^{\pi/3}8sec(x)csc(x)dx \ = \ 8\int_{\pi/6}^{\pi/3} \frac{dx}{sin(x)cos(x)}\)

\(\displaystyle Now, \ let \ u \ = \ \frac{sin(x)}{1+cos(x)} \ = \ tan(x/2), \ yields \ sin(x) \ = \ \frac{2u}{1+u^2},\)

\(\displaystyle cos(x) \ = \ \frac{1-u^2}{1+u^2}, \ and \ dx \ = \ \frac{2du}{1+u^2}.\)

\(\displaystyle Ergo, \ we \ have \ 8\int_{2-\sqrt3}^{\sqrt3/3}\frac{2du/(1+u^2)}{2u(1-u^2)/(1+u^2)^2} \ = \ 8\int_{2-\sqrt3}^{\sqrt3/3}\frac{1+u^2}{u(1-u^2)}du\)

\(\displaystyle = \ 8\int_{2-\sqrt3}^{\sqrt3/3}\bigg[\frac{1}{u}-\frac{1}{u-1}-\frac{1}{u+1}\bigg]du \ = \ ? \ \ The \ correct \ answer \ is \ 8ln(3).\)

\(\displaystyle Can \ you \ take \ it \ from \ here?\)

as you stated above \(\displaystyle \ The \ correct \ answer \ is \ 8ln(3).\)

i did get this answer as well.

 

[mmm4444bot]

Here's what I suspect, Ryan.

I'm thinking that you might have put the integral into software, getting something like:

ln(tan(b)) - ln(tan(a))

You then tried to reason backwards from this result.

If this is the case, then you didn't really get anything.

 

[BigGlenntheHeavy]

Note, Ryan for integrals involving rational functions of sine and cosine, the substitution

\(\displaystyle u \ = \ \frac{sin(x)}{1+cos(x)} \ = \ tan(x/2) \ will \ yield \ cos(x) \ = \ \frac{1-u^2}{1+u^2}, \ check;\)

\(\displaystyle u^2 \ = \ \frac{sin^2(x)}{(1+cos(x))^2} \ = \ \frac{1-cos^2(x)}{(1+cos(x))^2} \ = \ \frac{1-cos(x)}{1+cos(x)}\)

\(\displaystyle Solving \ for \ cos(x) \ in \ this \ equation, \ we \ have \ cos(x) \ = \ \frac{1-u^2}{1+u^2}\)

\(\displaystyle sin(x) \ = \ \frac{2u}{1+u^2}; \ u \ = \ \frac{sin(x)}{1+cos(x)}, \ sin(x) \ = \ u(1+cos(x)) \ = \ u\bigg(1+\frac{1-u^2}{1+u^2}\bigg) \ = \ \frac{2u}{1+u^2}\)

\(\displaystyle and \ dx \ = \ \frac{2du}{1+u^2}; \ u \ = \ tan(x/2) \ \implies \ \frac{x}{2} \ = \ arctan(u), \ dx \ = \ \frac{2du}{1+u^2}\)

\(\displaystyle Who \ originated \ this \ I \ don't \ know, \ perhaps \ it \ is \ lost \ in \ antiquity.\)

 

[Ryan Rigdon]

here is another way i did it and got the same answer of 8*ln(3)

 

[BigGlenntheHeavy]

\(\displaystyle As \ always \ Ryan \ there \ is \ more \ than \ one \ way \ to \ skin \ a \ cat.\)

\(\displaystyle You \ way \ is \ better \ (less \ grunt \ work), \ I \ missed \ it.\)

 

[Ryan Rigdon]

thanx for the compliment. they had an example in the book that used secxcscx = tanx + cotx, a trig identity i have never seen before. nice to learn something new.

 

[BigGlenntheHeavy]

\(\displaystyle Ryan, \ prove \ the \ identity \ sec(x)csc(x) \ = \ tan(x)+cot(x)\)

 

[Ryan Rigdon]

i cant prove it. i was just going by what the book told me is all.

 

[BigGlenntheHeavy]

\(\displaystyle You \ can't \ prove \ anything \ unless \ you \ try.\)