YAHT: Integral of the sqrt of tangent x

[MiataGirl]

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?

 

[Mill]

I would try seaching google. It would probably have the problem completely solved on a page.

 

[Yomicron]

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.

 

[goodoptics]

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.

 

[WinkOsmosis]

Wow! Integrals are the devil!

 

[oLLie]

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.

 

[GoodToGo]

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.

 

[Orsorum]

Probably substitute something else for y, then simplify that. If she has a table of forms, it'll have that, probably.

 

[notfred]

Originally posted by: Jellomancer
Wow! Integrals are the devil!

I'll agree with that.

 

[dabuddha]

but they can be your best friend too

 

[kherman]

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.

 

[Fandu]

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.