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Math Help - What is the integral of (y^.5)tanydy using integration by parts?

  1. #1
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    What is the integral of (y^.5)tanydy using integration by parts?

    What is the integral of y1/2tanydy? When I used integration by parts I get a recurring answer that I need to keep repeating integration by parts on.

    y1/2(-lncosy)-integral(.5y-.5)(-lncosy)

    Then I am stuck. i do not know how to integrate lncosy.
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    Re: What is the integral of (y^.5)tanydy using integration by parts?

    Are you sure you have the correct function? Also, is this an indefinite integral?
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    Re: What is the integral of (y^.5)tanydy using integration by parts?

    yes it is the correct function. i'm not sure if it is an indefinite integral or not. antiderivative of square root of y time tangent y dy
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    Re: What is the integral of (y^.5)tanydy using integration by parts?

    Sorry, but I don't know that this will be integrable. I could be wrong though so hopefully someone with more knowledge comes around.
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    Re: What is the integral of (y^.5)tanydy using integration by parts?

    Hey gp8283.

    Note that -ln(cosy) = ln(1/cosy) = ln(secy). You will need to do integration by parts for this but the derivative of ln(secy) = tan(y).

    u = tan(y), du/dy = 1/sec^2(x). y = arctan(u) which means y^(1/2) = SQRT(arctan(u)) tany*cos(y)/cos(y) = sin(y)cos(y)/cos^2(y) = sin(y)cos(y) * sec^2(y) so this integral changes to SQRT(arctan(u))*sin(arctan(u))*cos(arctan(u))*du.

    Now you can get formulas for sin(arctan(u)) and cos(arctan(u)) by using Pythagoras's theorem and the trig rations. sin(arctan(u)) = u/SQRT(1 + u^2) with cos(arctan(u)) = SQRT(1 - sin(arctan(u))^2). and y^(1/2) = SQRT(arctan(u)).

    I know this is a little out of the way, but the reason I mention this is because if you are solving a recurring integral (like say e^(x)*sin(x)) then you will get an expression somewhere that has the same as your starting integral and then collect both together and simplify.

    The integral you have mentioned above has y^(0.5) compared to y^(-0.5) which means that if you keep doing this you won't get the form that you can collect and simpify like the e^(x)*sin(x) example.

    Another way that would make things easier is to try the substitution y^(1/2) = tan(u) or even something like sin(u) or cos(u) since there are very specific formulas for getting trig functions in terms of arguments that have arc-sine, arc-cosine, or arc-tangent in them.
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    Re: What is the integral of (y^.5)tanydy using integration by parts?

    would it be possible to take ((((y^.5tanydy))^2))^.5 square the whole thing and take the square root of that to make y not to the power of 1/2, and just work with ytan^2ydy and then worry about the square root later on antidifferentiation using the chain rule after doing integration by parts on ytan^2dy?
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    Re: What is the integral of (y^.5)tanydy using integration by parts?

    To clarify, if g(y) = y*tan^2(y), then how are you going to calculate the integral in terms of Integral [g(y)]^(1/2)dy using the chain rule?
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