need help in the derivation of integration formula:

- anonymous

need help in the derivation of integration formula:

- chestercat

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- misty1212

HI!!

- misty1212

i didn't know there was such a formula

- anonymous

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

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## More answers

- anonymous

solve by taking x=a sect

- anonymous

no problem @misty1212

- misty1212

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

- misty1212

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

- misty1212

did you get to that part, or no?

- anonymous

yes

- anonymous

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

- misty1212

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

- anonymous

can u go after that

- misty1212

there is a "reduction formula" for secant

- misty1212

are you allowed to use it?

- anonymous

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.....

- anonymous

what is reduction formula

- misty1212

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

- anonymous

then what is that?

- misty1212

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

- misty1212

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

- anonymous

hav u done it by taking integration by parts?

- misty1212

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

- misty1212

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

- misty1212

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

- misty1212

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

- anonymous

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...

- misty1212

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"

- anonymous

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...

- misty1212

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

- misty1212

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

- ganeshie8

if you like partial fractions,
\[\sec x =\dfrac{\cos x}{\cos^2x} = \dfrac{\cos x}{1-\sin^2x}\]

- misty1212

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

- anonymous

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

- ganeshie8

same can be extended to \(\sec^3x\) too i think
\[\sec^3 x =\dfrac{\cos x}{\cos^4x} = \dfrac{\cos x}{(1-\sin^2x)^2}\]

- misty1212

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

- anonymous

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

- misty1212

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

- misty1212

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

- anonymous

oh yes i got that...

- ganeshie8

looks good to me !

- ganeshie8

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

- anonymous

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...

##### 2 Attachments

- anonymous

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

##### 1 Attachment

- anonymous

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