Integrating substitution problem?

[Gib Z]

Integration of Powers of Secant is not so difficult if you use a recursive formula. there's also less chance of error until the last few lines, since we haven't substituted any numbers for the pro numerals yet. It's not difficult to derive the formula, it takes about 2 minutes.

I know rocophysics will like this challenge, so I the only hint I give to start is let to use integration by parts =]

 

[Defennder]

It's quite tedious to integrate the cube of sec theta. My suggestion is that you first try to figure out how to integrate that as an indefinite integral then apply the result to evaluate the definite integral in the question. You'll definitely need to use integration by parts here, and if you do it correctly, you notice a point where you'll need to bring an integral on the RHS of the equation to to the LHS. You'll need at least 2 substitutions on this problem (just to integrate (sec(theta))^3 )

 

[Gib Z]

Once again someone ignores every other post in the thread >.<

 

[arildno]

Homework Statement

Sorry to hijack this thread sort of (as a similar named one already exists), but the title is aptly suited to my question.

I have integral to integrete and I don't really know how to do it tbh. . .

s=\int{\sqrt{2+(3t)^2}dt Limits are from t=0 to t=5 (how do I show this on latek?)

If it was 2-3t then I'd simply substitute the t for sin u, but what do I do when it positive?

I don't see why people bother with the tan substitution.

It is by far simplest to set:
t=\frac{\sqrt{2}}{3}Sinh(u)
whereby the integrand resolves itself to \frac{2}{3}Cosh^{2}(u)=\frac{3}{4}(Cosh(2u)+1)

 

[Gib Z]

Minor correction: The coefficient on the right hand side should be 1/3.

 

[rocomath]

Integration of Powers of Secant is not so difficult if you use a recursive formula. there's also less chance of error until the last few lines, since we haven't substituted any numbers for the pro numerals yet. It's not difficult to derive the formula, it takes about 2 minutes.

I know rocophysics will like this challenge, so I the only hint I give to start is let to use integration by parts =]

I'll give it a try after my finals :-]

Once again someone ignores every other post in the thread >.<

Never!

I don't see why people bother with the tan substitution.

It is by far simplest to set:
t=\frac{\sqrt{2}}{3}Sinh(u)
whereby the integrand resolves itself to \frac{2}{3}Cosh^{2}(u)=\frac{3}{4}(Cosh(2u)+1)

I actually do hyperbolic substitutions when I'm integrating, but I think it was far more beneficial to do it the long way here to learn how important the constant is and the usefulness of trig substitution.

 

[malty]

Hyperbolic functions eh?

Well to be honest, I'd would never even have consider them, learn something new everyday I guess.

To be honest though I much preferred the Tan method because it was fun (and annoying when I think of my integration by parts loop and not spotting the obvious i.e. Integral of Sec ^2 x equal Tan x) and it was also my first time attempting a secant to the power of .. integral.

Will also give a derivation proposed by Gibz a go over the holidays, no idea where to start though, but hopefully between now and then some sort of idea will have occurred.

 

[rocomath]

If you really get bored and want to become a Integrating Master (lol), do some of the problems in this thread ... I've only completed 5 of them since I don't have much spare time. Some make you just want to JUMP! :-]

How Good Am I? - https://www.physicsforums.com/showthread.php?t=149706