Question

need help in the derivation of integration formula:

Answer 1

HI!!

Answer 2

i didn't know there was such a formula

Answer 3

\[\int\limits_{.}^{.} \sqrt{x^{2}-a^{2}}dx =\frac{x }{ 2 }\sqrt{x ^{2}-a ^{2}}-\frac{ a ^{2} }{ 2 }\log \left| x+\sqrt{x ^{2}+a^2} \right|+C\]

Answer 4

solve by taking x=a sect

Answer 5

no problem @misty1212

Answer 6

this one is a pain in the neck i think, but the sub is right

Answer 7

you are going to get
\[a^2\int \sec^3(u)-\sec(u)du\]

Answer 8

did you get to that part, or no?

Answer 9

yes

Answer 10

i dont know what to do after that.....

Answer 11

this is the kind of thing you look up in the back of the text because it is boring beyond belief

Answer 12

can u go after that

Answer 13

there is a "reduction formula" for secant

Answer 14

are you allowed to use it?

Answer 15

u can take the snapshot for the things u solved in the notebook instead of typing them. it could be easier, if u can do it.....

Answer 16

what is reduction formula

Answer 17

you mean in general or "what is the reduction formula for \(\int \sec^n(x)dx\)?

Answer 18

then what is that?

Answer 19

here are a bunch of them
http://archives.math.utk.edu/visual.calculus/4/recursion.2/

Answer 20

in your case \(n=3\) so you get \[\int sec^3(u)du=\frac{1}{2}\tan(u)\sec(u)+\frac{1}{2}\int \sec(u)du\]

Answer 21

hav u done it by taking integration by parts?

Answer 22

it looks like that right? but it is not
it comes from doing some trig business and a u sub

Answer 23

like the same trick they use for \(\int\sin^n(x)dx\)
we can work through it if you like although it is not that interesting

Answer 24

oh to answer your question, no i used nothing but the"reduction formula" with \(n=3\)
the one i sent the link to

Answer 25

ignoring the annoying \(a^2\) out front, and combining like terms, we should be at \[\frac{a^2}{2}\sec(u)tan(u)-\frac{a^2}{2}\int \sec(u)du\]

Answer 26

i dont know these reduction formulas...it's not in my book....in my book, the derivation was given by the method of integration by parts and in the end it was written that u can take x=sec t to solve the problem...so i did it by doing the substitution and got clutched in the middle...

Answer 27

now how to integrate secant, again it is not interesting, best to memorize it
however, you want a good explanation, easier than i can write here, click on this
http://math2.org/math/integrals/tableof.htm
then go to "proof"

Answer 28

i can prove by the method of integration by parts...so no problem, when i would learn about these reduction formulas then i would go by that method also...

Answer 29

i think you use parts for \(\int \sec^3(x)dx\) but not for \(\int \sec(x)dx\)

Answer 30

it is just something they do to make a formula is all
some people like formulas

Answer 31

if you like partial fractions,

\[\sec x =\dfrac{\cos x}{\cos^2x} = \dfrac{\cos x}{1-\sin^2x}\]

Answer 32

oh cool, lots easier than the un-intuitive multiplying top and bottom business

Answer 33

please give a more detail where to start from....@ganeshie8

Answer 34

same can be extended to \(\sec^3x\) too i think

\[\sec^3 x =\dfrac{\cos x}{\cos^4x} = \dfrac{\cos x}{(1-\sin^2x)^2}\]

Answer 35

@ganeshie8 is proving \[\int\sec(x)dx=\ln(\sec(x)+\tan(x))\]

Answer 36

i dont know how to integrate that partial fraction....@ganeshie8

Answer 37

\[\int \frac{\cos(x)}{1-\sin^2(x)}dx\] put \(u=\sin(x)\) and integrate \[-\int \frac{du}{1-u^2}\] using partial fractions

Answer 38

@ganeshie8 that is correct yes?
never saw it done this way

Answer 39

oh yes i got that...

Answer 40

looks good to me !

Answer 41

for \(\int\sec^3x\,dx\) i would try reduction formula though as partial fractions looks a bit lengty

http://www.wolframalpha.com/input/?i=%5Cint+1%2F%281-u%5E2%29%5E2

Answer 42

the problem is that i dont know these reduction formulas and the link that @ganeshie8 has given....i am a beginner in integration i only know the formulas given in my book....i wold rather use integration by parts from the beginning than using all such formulas...that is easiest...

Answer 43

the statement in the middle had initiated me...@ganeshie8

Answer 44

@ganeshie8