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
A definite trigonometric integral is a type of integral in calculus that involves trigonometric functions, such as sine, cosine, and tangent. It is used to find the area under a curve or the value of a function at a specific interval.
To solve a definite trigonometric integral, you can use various integration techniques, such as substitution, integration by parts, or trigonometric identities. It is important to first identify the type of integral and then choose the appropriate method for solving it.
Some of the common trigonometric identities used in definite trigonometric integrals include the Pythagorean identities, double angle identities, and half angle identities. These identities help to simplify the integral and make it easier to solve.
Yes, definite trigonometric integrals can be solved using a calculator, but it is important to make sure the calculator is set to the correct mode (degrees or radians) and to double-check the answer for accuracy.
Definite trigonometric integrals have many real-life applications, such as in physics, engineering, and astronomy. They are used to calculate the displacement, velocity, and acceleration of moving objects, and to analyze the behavior of waves and periodic phenomena.