Integration question from terry lee :( (1 Viewer)
- Thread starter coeyz
- Start date
gurmies
Drover
Dang, beat me to it. Although the second question has a whole square root involved.
Last edited:
gurmies
Drover
Last edited:
edit: yea my bad, didn't see square root
\int \sqrt{\frac{x-1}{x+1}}dx
\\=\int \frac{x-1}{\sqrt {(x-1)(x+1)}}dx
\\=\int \frac{x-1}{\sqrt {x^2 -1}}dx
\\=\int \frac{x}{\sqrt {x^2 -1}}dx - \int \frac{1}{\sqrt {x^2 -1}}dx
\\=\frac{1}{2}\int (x^2-1)^{-\frac{1}{2}}\frac{d}{dx}(x^2-1)dx - ln( x + \sqrt {x^2 -1})
\\= \sqrt {x^2 -1} - ln( x + \sqrt {x^2 -1}) + C )
for the first question, wht happen if the limit is 0 to 4?
coz when u sub it in i cant get the answer ;(
the answer is -4 square root 3 + 3 ln ( 5+3sqrt3 / 1+3sqrt3)
thanks againn
gurmies
Drover
=) woot
oh no cant do itoh can do it now~
thankss
i do it up to:
3/sqrt2 [ tan^-1 5/sqrt2 - tan^-1 1/sqrt2] - ln9
anyone please help :'(
gurmies
Drover
you sure that's the answer? I get ln (1/9) ...
ye the ans is -4 square root 3 + 3 ln ( 5+3sqrt3 / 1+3sqrt3)
Could the answer be ??
which is what you worked out, coeyz
okayy i think the answer is wrong dw
thanks a lot anyway
how did u get the middle part?Could the answer be ??
which is what you worked out, coeyz
what is [ tan^-1 (5/sqrt2) - tan^-1 (1/sqrt2) ] ?
GUSSSSSSSSSSSSS
Active Member
its just one where u have the derivative on top and the function on the bottom
and then the 2 and a half cancel out in the process
....lol i explained that real bad...sorry!
Let phi = tan^-1 (5/sqr(2) ) - tan^-1(1/sqr(2))
Therefore: tan(phi) = [tan (tan^-1(5/(sqr2) - tan(tan^-1(1/sqr2)] / [ 1 + tan (tan^-1(5/sqr(2)) ) x tan(tan^-1(1/sqr 2))]
= [5/sqr 2 - 1 / sqr 2] / [ 1 + 5/sqr 2 x 1/sqr 2 ]
= 4/sqr 2 / (7/2) = 4 sqr 2 / 7
Therefore phi = tan^-1 (4 sqr 2)/7
gurmies
Drover
^ lolwut
LOL! Terry Lee also happens to be a guy in coeyz and my grade. Who unpermed his hair. He looked so cool before!