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Math Help - Another inverse sine integral check

  1. #1
    Newbie alysha27's Avatar
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    Integral by Parts

    Hi, If anyone could check the solution I got to this problem, that'd be great.

    Q: The integral of xsin^-1(x) dx

    My solution: To solve, I first used integration by parts, and then with the integral left over by parts ( of x squared over 2 times the square root of one minus x squared) I used trigonometric substitution to solve the rest.

    The answer I got was:

    1/2*x^2*sin^-1(x) - 1/4*sin^-1(x) + 1/4*x*(1-x^2)^1/2 +C


    Thanks a lot for the help, I appreciate it.
    Last edited by alysha27; February 26th 2006 at 06:20 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by alysha27
    Hi, If anyone could check the solution I got to this problem, that'd be great.

    Q: The integral of xsin^-1(x) dx

    My solution: To solve, I first used integration by parts, and then with the integral left over by parts ( of x squared over 2 times the square root of one minus x squared) I used trigonometric substitution to solve the rest.

    The answer I got was:

    1/2*x^2*sin^-1(x) - 1/4*sin^-1(x) + 1/4*x*(1-x^2)^1/2 +C


    Thanks a lot for the help, I appreciate it.
    That is correct.

    RonL
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  3. #3
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    Quote Originally Posted by alysha27
    Hi, If anyone could check the solution I got to this problem, that'd be great.

    Q: The integral of xsin^-1(x) dx

    My solution: To solve, I first used integration by parts, and then with the integral left over by parts ( of x squared over 2 times the square root of one minus x squared) I used trigonometric substitution to solve the rest.

    The answer I got was:

    1/2*x^2*sin^-1(x) - 1/4*sin^-1(x) + 1/4*x*(1-x^2)^1/2 +C


    Thanks a lot for the help, I appreciate it.
    The easiest way is to take the derivative of the anti-derivative and see if it returns back the function.
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