Applied Calculus Help
Hi,
Can anyone explain how to integrate (1-x^2)^1/2 using integration by parts as a solution I looked at is pretty confusing?
Thanks
Can anyone explain how to integrate (1-x^2)^1/2 using integration by parts as a solution I looked at is pretty confusing?
Thanks
Original post by theoneandonlyyo
Hi,
Can anyone explain how to integrate (1-x^2)^1/2 using integration by parts as a solution I looked at is pretty confusing?
Thanks
Can anyone explain how to integrate (1-x^2)^1/2 using integration by parts as a solution I looked at is pretty confusing?
Thanks
Why by parts? The usual (famous) way is with a trig substitution. Or a sketch and a bit of geometry.
(edited 1 year ago)
Original post by mqb2766
Why by parts? The usual (famous) way is with a trig substitution. Or a sketch and a bit of geometry.
Idk my lecturer wants us to know this way too
Original post by theoneandonlyyo
Idk my lecturer wants us to know this way too
You said youve got a solution? Can you upload it.
As far as I can see, you can do by parts but then you have to do the "same" trig substitution that you could do straight away. So kinda pointless.
(edited 1 year ago)
Original post by theoneandonlyyo
Idk my lecturer wants us to know this way too
Just for completeness, a quick google this morning gave
https://math.stackexchange.com/questions/533082/integral-of-sqrt1-x2-using-integration-by-parts
and the multiplication by 1, then by parts, then trig substitution/spot integral was the way I was imagining last night. Whether you argue that spotting you can get a standard integral for arcsin is not doing trig substitution is debatable. You know the solution must contain an arcsin as the sector area is based on angle but the integral limits are based on side lengths/ratios so there must be an inverse trig kicking around in there somewhere.
A bit of head scratching this morning gave another by parts by doing dots on the original integrand then by parts on the two terms but you end up doing a more complex trig substitution at the end.
(edited 1 year ago)
Original post by mqb2766
Just for completeness, a quick google this morning gave
https://math.stackexchange.com/questions/533082/integral-of-sqrt1-x2-using-integration-by-parts
and the multiplication by 1, then by parts, then trig substitution/spot integral was the way I was imagining last night. Whether you argue that spotting you can get a standard integral for arcsin is not doing trig substitution is debatable. You know the solution must contain an arcsin as the sector area is based on angle but the integral limits are based on side lengths/ratios so there must be an inverse trig kicking around in there somewhere.
A bit of head scratching this morning gave another by parts by doing dots on the original integrand then by parts on the two terms but you end up doing a more complex trig substitution at the end.
https://math.stackexchange.com/questions/533082/integral-of-sqrt1-x2-using-integration-by-parts
and the multiplication by 1, then by parts, then trig substitution/spot integral was the way I was imagining last night. Whether you argue that spotting you can get a standard integral for arcsin is not doing trig substitution is debatable. You know the solution must contain an arcsin as the sector area is based on angle but the integral limits are based on side lengths/ratios so there must be an inverse trig kicking around in there somewhere.
A bit of head scratching this morning gave another by parts by doing dots on the original integrand then by parts on the two terms but you end up doing a more complex trig substitution at the end.
Thank you so much for this solution! I am still very confused tho haha but I will just stick to the trig substitution method. It won't let me upload the solution too:/
Original post by theoneandonlyyo
Thank you so much for this solution! I am still very confused tho haha but I will just stick to the trig substitution method. It won't let me upload the solution too:/
WHat are you confused about (in the stackexchange solution)?
Original post by mqb2766
WHat are you confused about (in the stackexchange solution)?
The I bit
Original post by theoneandonlyyo
The I bit
Would help to be a bit more specific, but if youre referring to the integral of
-x^2/sqrt(1-x^2)
then adding zero (1-1) to the numerator gives two integrals
(1-x^2)/sqrt(1-x^2)
and
-1/sqrt(1-x^2)
The latter is the usual arcsin() integral and the former is the original integral, so I. So you can take it over to the left hand side and divide by 2 to get the result.
If you cant see it, write up (and post) what you do understand.
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