The discussion revolves around evaluating the definite integral ##\int_0^1 x^6 \arcsin{x} dx##, which involves trigonometric functions and integration techniques.
There is an ongoing exploration of different approaches to the integral. Some participants have successfully used integration by parts and are questioning whether there are alternative methods that could simplify the process. The conversation reflects a mix of shared experiences and suggestions for further exploration.
Participants note the complexity of the integral and the potential for using definite integral properties, while also acknowledging the tedious nature of multiple integration by parts. There is mention of rough work and standard substitutions that may have been used in attempts to solve the integral.
PS- I solved it using Integration by parts. Looking for a shorter method if existsKrushnaraj Pandya said:Homework 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 methodnomadreid said:Try integrating by parts several times. That is a little tedious but will get you the final answer.
Krushnaraj Pandya said: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)nomadreid said: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)Ray Vickson said: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 simplificationnomadreid said:Sounds 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 thisKrushnaraj Pandya said: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 simplification