Integration by Parts: Solving Homework

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The discussion focuses on solving the integral of the inverse sine function, specifically \(\int \sin^{-1}(x) \, dx\). Participants clarify the correct notation for the inverse sine function, emphasizing that it should be denoted as \(\arcsin(x)\). The integration by parts formula is discussed, with some confusion regarding the variables used. Ultimately, the correct selections for \(u\) and \(dv\) are confirmed, leading to the solution: \(x \cdot \arcsin(x) + \sqrt{1-x^2} + C\). The thread concludes with the original poster expressing gratitude for the assistance received.
crm08
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Homework Statement



\int(sin(x)^{-1}), dx

Homework Equations



*By Parts Formula: f(x)g(x) - \int(g(x) f'(x)) dx

Also for d/dx sin(x)^{-1} I used 1/sqrt(1-x^{2})

The Attempt at a Solution



Just started learning this method, I tried letting f(x) = sin(x)^{-1} and g(x) = dx but nothing really simplified, can someone help with selecting the correct g(x) and f(x), Thanks
 
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crm08 said:

Homework Statement



\int(sin(x)^{-1}), dx

Is the function you're trying to integrate here the inverse function of sin(x)? If so, it should be denoted \arcsin(x)=\sin^{-1}(x)

Homework Equations



*By Parts Formula: f(x)g(x) - \int(g(x) f'(x)) dx

This isn't a formula, since you haven't specified what it is equal to! The formula I would use is \int v du=uv-\int udv. Is this the formula you have been taught? If not, what is the formula you have been taught?
 
Yes, the problem is asking for the integral of arcsin(x), and also yes, that is the formula we are using, my "u's" and "v's" look a lot alike sometimes so I replaced them with f(x) and g(x), sorry about the confusion
 
crm08 said:
Yes, the problem is asking for the integral of arcsin(x), and also yes, that is the formula we are using, my "u's" and "v's" look a lot alike sometimes so I replaced them with f(x) and g(x), sorry about the confusion

Ok, so your selection was u=\sin^{-1}(x) \,\, , \,\, dv=dx, right? So, what went wrong? This is the choice that I would make!
 
Ok nevermind I got it now, I was working towards an answer to this problem that my 89 gave me but I typed it in wrong, I see how to to it now, the answer being:

x*arcsin(x) + sqrt(1-x^2)

Thanks for your help
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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