Are you sure you have the correct function? Also, is this an indefinite integral?
What is the integral of y^{1/2}tanydy? When I used integration by parts I get a recurring answer that I need to keep repeating integration by parts on.
y^{1/2}(-lncosy)-integral(.5y^{-.5})(-lncosy)
Then I am stuck. i do not know how to integrate lncosy.
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.
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?