Great vid, but got a question because I've seen another answer: x(arcsinx)^2+2*arcsinx*cos(arcsinx)-2x+c. So instead of SQRT{1-x^2}, they use cos(arcsinx) in the middle of the answer. Does anybody know whats up with that?
Imagine a right angled triangle with an angle a. The sin value of that angle is x. Therefore, the sin value can be expressed as x/1 (opposite over hypothenuse). To figure out cos(a), pythagoras theorem can be used to figure out the adjacent side of the angle which can be expressed as sqrt(1-x^2), which is the cosine of that since the hypothenuse is 1. So, cos(arcsinx) can be simplified into cos(a) which is the same as sqrt(1-x^2). So that means that sqrt(1-x^2) and cos(arcsinx) are the same thing
It comes from using U substitution after the first integration by parts. Letting u = arcsin x means that sin u = x. Since the derivative of arcsinx (1/(sqrt(1-x^2)) is in the integral, after u sub u are left with an easy integral of u sinu which can easily be solved with integration by parts and then arcsin x is subbed back in after.
That's good speaking and details perform logic process. but that taken long process and twice IBP spent 4 minutes. I think of you are good calculus lecturer. I have a Smart Formula for this integral, that give up IBP and just one step by one smart Formula that's 5 seconds. if you interesting in this, I will tell you late.
What an absolutely heinous integral, but now I've learned, don't give up on your u choice until at least 2 rounds of integration by parts!
ur explanations are always so clear thank u
thank you!!
Thank you this was extremely helpful, took me forever to understand the second integration but you explained it incredibly well.
No i didnt knew that
Good video. Were the results of the reduction formulas used in this procedure?
ty man you save me
You are welcome!
Great vid, but got a question because I've seen another answer: x(arcsinx)^2+2*arcsinx*cos(arcsinx)-2x+c. So instead of SQRT{1-x^2}, they use cos(arcsinx) in the middle of the answer. Does anybody know whats up with that?
Imagine a right angled triangle with an angle a. The sin value of that angle is x. Therefore, the sin value can be expressed as x/1 (opposite over hypothenuse). To figure out cos(a), pythagoras theorem can be used to figure out the adjacent side of the angle which can be expressed as sqrt(1-x^2), which is the cosine of that since the hypothenuse is 1. So, cos(arcsinx) can be simplified into cos(a) which is the same as sqrt(1-x^2). So that means that sqrt(1-x^2) and cos(arcsinx) are the same thing
It comes from using U substitution after the first integration by parts. Letting u = arcsin x means that sin u = x. Since the derivative of arcsinx (1/(sqrt(1-x^2)) is in the integral, after u sub u are left with an easy integral of u sinu which can easily be solved with integration by parts and then arcsin x is subbed back in after.
tysm
I knew that
If you want to solve this integral, with different way. Contact me
now who wants to know the intégral of this 😞
Thank you
You are welcome!
That's good speaking and details perform logic process. but that taken long process and twice IBP spent 4 minutes. I think of you are good calculus lecturer. I have a Smart Formula for this integral, that give up IBP and just one step by one smart Formula that's 5 seconds. if you interesting in this, I will tell you late.
thank you
you are welcome:)
Oh my god.