Definite trigonometric integral

[Krushnaraj Pandya]

Homework Statement

solve $\int_0^1 x^6 \arcsin{x} dx$

 

[Krushnaraj Pandya]

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

PS- I solved it using Integration by parts. Looking for a shorter method if exists

 

[nomadreid]

Try integrating by parts several times. That is a little tedious but will get you the final answer.
Edit: wrote the above before your PS came up. I think by parts will be the shortest way.

 

[Krushnaraj Pandya]

Try 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

 

[nomadreid]

did you get (35π -32)/490 ?

 

[Ray Vickson]

I solved it by a single integration by parts...I'm looking for an even shorter method

Show your work. Maybe what you did is the shortest possible way, but we cannot tell if you don't show us.

 

[Krushnaraj Pandya]

did you get (35π -32)/490 ?

Yes I did. (I took sin inverse as first function and x^6 as second, after 2 or 3 substitutions I got this answer)

 

[Krushnaraj Pandya]

Show your work. Maybe what you did is the shortest possible way, but we cannot tell if you don't show us.

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)

 

[nomadreid]

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.

 

[Krushnaraj Pandya]

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.

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

 

[Krushnaraj Pandya]

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

this is probably what the question-maker had in mind I guess, I see no other way to shorten or quicken this