Integration problem

Philip K.

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Apr 28, 2020
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May someone help me to solve the following integral :

∫ {0}^{1} (arcsin(x))^2 dx

I know i must first solve the indefinite integral using intgration by parts rule

∫(arcsin(x))^2 dx

I usually know how to apply integration by parts, however here i see only one variable but i need 2 for the rule.
I am stick in the beginning. I just dont see how i may apply intgration by parts.
 
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May someone help me to solve the following integral :

∫ {0}^{1} (arcsin(x))^2 dx

I know i must first solve the indefinite integral using intgration by parts rule

∫(arcsin(x))^2 dx

I usually know how to apply integration by parts, however here i see only one variable but i need 2 for the rule.
I am stick in the beginning. I just dont see how i may apply intgration by parts.
If I were to do this problem, I would first use substitution:

arcsin(x) = Θ

x = sin(Θ)........................... continue............
 
No, what you are saying is not correct. The two pieces of the integrand do not have to have "two variables" to use integration by parts!
One of the two factors you are looking for could be 1 !! Other times you can even introduce a true function of x (not a constant) and then multiply by the reciprocal.

A classic integration by parts problem is [math] \int ln(x)dx[/math]
 
Another point I want to make. Even if you thought that to use integration by parts you needed to functions of x (which is not true), then why not break up (arcsin(x))^2 into (arcsin(x))*(arcsin(x)).

More importantly why do you think that this problem needs to be done by parts?
 
No, what you are saying is not correct. The two pieces of the integrand do not have to have "two variables" to use integration by parts!
One of the two factors you are looking for could be 1 !! Other times you can even introduce a true function of x (not a constant) and then multiply by the reciprocal.

A classic integration by parts problem is [math] \int ln(x)dx[/math]

Yes, that was my mistake. I was looking for second variable.
Now i manage to solve it, thank you.
 
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