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YAHT: Integral of the sqrt of tangent x

MiataGirl

Banned
this isn't for me btw..i don't even know what that means 🙂

but my friend says that the solution is about a 1/4 page long, and her ti-89 took about 3 minutes to compute it. i told her i knew some people that might help.

supposedly it's a rather popular problem too? basically she just wants to know how to get started...how to continue..and possible how to finish? 🙂
 
the solution is huge, Maple even hesitated solving it

here you go,
1/2*tan(x)^(1/2)/(cos(x)*sin(x))^(1/2)*cos(x)*2^(1/2)*arccos(cos(x)-sin(x))-1/2*sqrt(2)*ln(cos(x)+sqrt(2)*sqrt(tan(x))*cos(x)+sin(x))

are you sure your friend is reading the problem correctly? it seems kinda complicated for a homework assignment

EDIT: no wait... Maple didn't hesitate at all, it was just some process in the background slowing down my computer.
 
There should be a table of integrals in the back of your friend's Calculus book. The one your friend needs should be listed under Trigonometric forms.
 
There's a rule for integration of tan to a power, I'm assuming as long as it's not 1 it might work? It can't be 1 because then you'd divide by one. I'm too lazy to type it in here too.
 
Try this:
Substitute tan(x) = y
dy = sec^2(x)
Mutiply and divide the integral by sec^2(x). Now theres one more identity to remember:
1+tan^2(x) = sec^2(x).
So basically the integral changes to sqrt(y)/(1+y^4). However I have no idea how to solve this 😀 Its been a while. But I am sure your friend can try and solve this.
 
Probably substitute something else for y, then simplify that. If she has a table of forms, it'll have that, probably.
 
PRAISE ALLAH!!!
I no longer need to know how to do this and I am fanatical about the fact!!!
I used this stuff in calculus, school and ummm .... .that's it. Brain feals much better these days.
 
haha, HP's still better than TI! My 49G had the answer in under 15 seconds. 🙂

The answer is pretty long...... I don't really want to type all that out. But I don't think it's actually that hard, GoodToGo is on the right track. If you can solve the integral that he posted directly, then you'll have the answer. But I would try making an inverse trig substitution, and then using integration by parts. I might have to try and work that one by hand a bit later tonight.
 
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