Came across this gem while studying

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While studying for my upcoming Cal B I plugged one problem I was having problems with into Wolfram's Integral Calculator. I got an unexpectedly long answer... does Wolfram often do this or was I just lucky?

Please tell me this isn't one equation...

Yes, yes it is. It is the integral of (x^5)(sec x)... according to Wolfram. Coincidentally, I dont know hot to integrate that properly. Any ideas WP?

This isn't one equation. I only wrote this because you said please.

ruveyn

http://integrals.wolfram.com/index.jsp?expr=5^(x)+Sec[x]&random=false


HOW THE HELL DOES IT WORK!?
I need ask my old math teacher

The following way may be used:

∫(u (x) v'(x)) dx = u (x) v (x) - ∫(u' (x) v(x)) dx

u (x) := x^5
v' (x) := sec (x)

so we need v (x):

v (x) = ∫sec (x) = ∫(1/cos (x))dx

according to Bronshtein, 1.1.3.3, #327:

v (x) = ln (tan (x/2 + π/4))

so we can say:

∫(x^5 sec (x)) dx = x^5 * (ln (tan (x/2 + π/4))) - ∫((x^4)/6 * (ln (tan (x/2 + π/4)))) dx

Not to make more complicated we define:

X_1 = x^5 * (ln (tan (x/2 + π/4)))
X_2 = ((x^4)/6 * (ln (tan (x/2 + π/4))))

and have:

∫(x^5 sec (x)) dx = X_1 + ∫X_2 dx

To solve

∫X_2 dx

we need to redefine u and v:

u (x) := (x^4)/6
v' (x) := (ln (tan (x/2 + π/4)))) dx

If you make a substitution of

w := (x/2 + π/4)

with

dw/dx = 1/2
dx = 2dw

you have

v (x) = ∫(v'(x)) dx = ∫(ln (tan (w)) dx = ∫2(ln (tan (w)) dw

and

s := tan (w)
ds/dw = - ln (cos s)
dw = (-1)/(ln (cos s))

you have

v (x) = ∫(-2) ln (s)/(ln (cos (s))) ds

if you transform this into a form of

∫(g (x) (f'(x)/f(x)))dx

with

∫(f'(x)/f(x))dx = ln |f(x)|

the term X_2 shall be able to solve quite simply. The rest is on you, but I am quite certain that the solution is a lesser monster than the automatically created wolfram-version.

Last edited by Dussel on 26 Jan 2009, 6:14 am, edited 1 time in total.

How does one determine if that monstrosity is even correct. It looks like something Ramanujan would have jotted down on a rainy afternoon in England.

ruveyn

sometimes I just don't know when to keep my mouth shut. I'm just a hardware jockey, I never got past Algebra 2...

Wow. That equation looks fun! Put that on a test and see the teacher go crazy! lol It looks like Wolfram gave out Fourier transforms with complex variables? Then again, I'm just starting to relearn math again. I most likely am wrong.

Same here, I never got past Algebra 1...Yet somehow I graduated.

I could not stop myself on working further on this "little nut" and it is really not trivial to find an elegant solution.

it is a matter of integration by parts. You can only integrate it one way (by reducing the x^5) because the antiderivative of sec... still includes sec. So your answer will keep getting more and more complicated until the x^5 is gone.

my chinese math professor would give insane problems like that and tell us to find a "tricky" and if we couldn't solve it in 3 sentences or less it was wrong.

when I look at that I keep thinking that there is a tricky to make it easier.


like integrating e^x*sinx something.

I think in the meanwhile the idea to go via

∫(u (x) v'(x)) dx = u (x) v (x) - ∫(u' (x) v(x)) dx

is too obvious, I play around more on the line of

∫(g (x) (f'(x)/f(x)))dx

with

∫(f'(x)/f(x))dx = ln |f(x)|




Could be

you could also play with trig identities or still use
∫(u (x) v'(x)) dx = u (x) v (x) - ∫(u' (x) v(x)) dx but with nontraditional u and v subs.

I played around with this integral last night and I found a quite straight forward way to solve it;

so we have:



so we can write further:



We define further:


For both integral Bronstein shows solution in integral No. 470 and 473. So it is less tricky than originally thought