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Math Help - integration of area of inverse trig

  1. #1
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    integration of area of inverse trig

    find the area between the curve y=tan-1x, the X axis and the line x=sqrt3

    find the area of the region enclosed by the curves
    y=2sin[(pi)x/4] and x=2sin[(pi)y/4]

    thankyou =]
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  2. #2
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    Quote Originally Posted by christina View Post
    find the area between the curve y=tan-1x, the X axis and the line x=sqrt3
    This is, of course, \int_{x=0}^{\sqrt{3}} tan^{-1}(x) dx. To do that integral, use integration by parts with u= tan^{-1}(x), dv= dx.

    find the area of the region enclosed by the curves
    y=2sin[(pi)x/4] and x=2sin[(pi)y/4]
    These two curves intersect at (0,0) and (2,2) so this is [tex]\int_{x=0}^2 (2sin(\pi x/4)- 4 sin^{-1}(x/2)/4) dx.
    You should be able to look up the integrals of both those functions.

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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by christina View Post
    find the area between the curve y=tan-1x, the X axis and the line x=sqrt3

    find the area of the region enclosed by the curves
    y=2sin[(pi)x/4] and x=2sin[(pi)y/4]

    thankyou =]
    For #1

    Graphical hint

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