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Math Help - integration by parts/ inverse trig function

  1. #1
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    integration by parts/ inverse trig function

    here is the problem
    evaluate the Integral
    arcsin(x)/ sqrt(1+x^2)
    the limits are 0 and .5

    I've used integration by part and came up with the equation of
    integral = arcsin^2(x) - arctanx
    then i put in the limits and got 873.44 this is when the calc is set to degrees
    but i have a feeling im doing this wrong
    please i need help i have a mid term later today


    here is some of my work
    u= arcsinx dv=1/sqrt(1+x^2)
    du= 1/sqrt(1+x^2) v=arcsinx
    arcsinx/ sqrt(1+x^2)= arcsinx(arcsinx)- intg (1/sqrt(1+x^2)^2

    this is then simplified to
    arcsinx^2-arctanx
    then by plugging in .5
    it becomes (pi/6)^2- arctan(.5)
    which comes out to -.189
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Pennsylvania
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    Quote Originally Posted by Lone21 View Post
    here is the problem
    evaluate the Integral
    arcsin(x)/ sqrt(1+x^2)
    the limits are 0 and .5

    I've used integration by part and came up with the equation of
    integral = arcsin^2(x) - arctanx
    then i put in the limits and got 873.44 this is when the calc is set to degrees
    but i have a feeling im doing this wrong
    please i need help i have a mid term later today


    here is some of my work
    u= arcsinx dv=1/sqrt(1+x^2)
    du= 1/sqrt(1+x^2) v=arcsinx
    arcsinx/ sqrt(1+x^2)= arcsinx(arcsinx)- intg (1/sqrt(1+x^2)^2

    this is then simplified to
    arcsinx^2-arctanx
    then by plugging in .5
    it becomes (pi/6)^2- arctan(.5)
    which comes out to -.189
    If this is indeed your integral \int_0^{\frac{1}{2}}\frac{\arcsin(x)}{\sqrt{1+x^2}  }, then you cannot do this by elementary terms. Are you sure it wasn't

    \int_0^{\frac{1}{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    If this is indeed your integral \int_0^{\frac{1}{2}}\frac{\arcsin(x)}{\sqrt{1+x^2}  }, then you cannot do this by elementary terms. Are you sure it wasn't

    \int_0^{\frac{1}{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}
    yes thats the correct question
    i think i was on the right path but just had the wrong equation
    if you could evaluate how its done that would be great
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Lone21 View Post
    yes thats the correct question
    i think i was on the right path but just had the wrong equation
    if you could evaluate how its done that would be great
    Do you still agree with your work? \frac{d}{dx}\bigg[\arcsin(x)\bigg]=\frac{1}{\sqrt{1{\color{red}-}x^2}}
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