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Krushnaraj Pandya
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Homework Statement
solve ##\int_0^1 x^6 \arcsin{x} dx##
Last edited:
PS- I solved it using Integration by parts. Looking for a shorter method if existsHomework Statement
solve ##\int_0^1 x^6 \arcsin{x} dx##
3. The Attempt at a Solution
I tried substituting arcsinx=y, but abandoned it a few steps later ahead as it started taking a more complicated form. I have a feeling definite integral properties play a larger role here and tried applying some but didn't get anywhere. I'd appreciate some help, thanks
I solved it by a single integration by parts...I'm looking for an even shorter methodTry integrating by parts several times. That is a little tedious but will get you the final answer.
I solved it by a single integration by parts...I'm looking for an even shorter method
Yes I did. (I took sin inverse as first function and x^6 as second, after 2 or 3 substitutions I got this answer)did you get (35π -32)/490 ?
I simply took arcsinx as first function, and x^6 as second function. It was just a simple by-parts and the second term (derivative of arcsin*integral of x^6) was an algebraic term- I just used 2 substitutions to evaluate that and get the second term as well. What I mean by a shorter method is whether I can avoid the entire by-parts procedure and just use definite-integral properties to calculate this (i.e by a different approach)Show your work. Maybe what you did is the shortest possible way, but we cannot tell if you don't show us.
haha, I guess a direct by parts is the shortest method possible. keeping arcsin as arcsin seems a better option, as for the substitutions I used- Its totally lost somewhere in my ton of rough work (side effects of a calculus course) but I remember It was something pretty standard like setting the term in the square-root in the denominator as t^2 and then further simplificationSounds good. Which substitutions?
PS (after your edit): the parts method is pretty straightforward. If you want to play around a bit, you can also express arcsin x as -i*ln(i*x + (1-x2)½), but that would be longer to work with, and you would still end up using parts.
this is probably what the question-maker had in mind I guess, I see no other way to shorten or quicken thishaha, I guess a direct by parts is the shortest method possible. keeping arcsin as arcsin seems a better option, as for the substitutions I used- Its totally lost somewhere in my ton of rough work (side effects of a calculus course) but I remember It was something pretty standard like setting the term in the square-root in the denominator as t^2 and then further simplification