Reduction formula trouble?

H

HappyCalculusStudent

New member
Joined
Jun 8, 2009
Messages
12
The other day I was having trouble calculating the definite integral
?(sec(t/2))^4 dt (from 0 to ?/2)
Thank you very much for your help! Using substitution, I get the correct answer of 8/3 .

Now the problem is that I'm not comfortable applying the formulas from tables, and I still don't understand is why I get other (wrong) answers. For instance, when using a table of integrals - reduction formula for ?(sec(u))^n du
I get two other answers, below:

Given the reduction formula
?(sec(u))^n du = 1/(n-1) tan(u)(sec(u))^(n-2) + (n-2)/(n-1)?(sec(u))^(n-2) du
I let n=4 and u=(t/2)

1/3 tan(u)(sec(u))^2 + 2/3?(sec(u))^2 du
changing the limits of integration??? 0 to ?/4
and substituting, I get
1/3 tan(?/4)(sec(?/4))^2 + 2/3 tan (?/4)
=1/3 (1)(2) + 2/3 = 2/3 + 2/3 = 4/3

The other time I did the first part the same.
I let n=4 and u=(t/2)
1/3 tan(u)(sec(u))^2 + 2/3?(sec(u))^2 du
but I calculated the latter integral, substituting u=t/2.
Then 2du=dt and
1/3 tan(t/2)(sec(t/2))^2 + 2/3*2(tan(t/2)
evaluating from 0 to?/2
=1/3 (1)(2) + 4/3 = 2/3 + 4/3 = 2

Q: What am I doing wrong?
This is bothering me. I'd like to learn to use the tables properly when u is a fraction or multiple of the variable. I'd really appreciate your advice.
 
HappyCalculusStudent said:
The other day I was having trouble calculating the definite integral
?(sec(t/2))^4 dt (from 0 to ?/2)
Thank you very much for your help! Using substitution, I get the correct answer of 8/3 .

Now the problem is that I'm not comfortable applying the formulas from tables, and I still don't understand is why I get other (wrong) answers. For instance, when using a table of integrals - reduction formula for ?(sec(u))^n du
I get two other answers, below:

Given the reduction formula
?(sec(u))^n du = 1/(n-1) tan(u)(sec(u))^(n-2) + (n-2)/(n-1)?(sec(u))^(n-2) du
I let n=4 and u=(t/2)

and your integral becomes

2?(sec(u))^4 du (from 0 to ?/4)



1/3 tan(u)(sec(u))^2 + 2/3?(sec(u))^2 du
changing the limits of integration??? 0 to ?/4
and substituting, I get
1/3 tan(?/4)(sec(?/4))^2 + 2/3 tan (?/4)
=1/3 (1)(2) + 2/3 = 2/3 + 2/3 = 4/3

The other time I did the first part the same.
I let n=4 and u=(t/2)
1/3 tan(u)(sec(u))^2 + 2/3?(sec(u))^2 du
but I calculated the latter integral, substituting u=t/2.
Then 2du=dt and
1/3 tan(t/2)(sec(t/2))^2 + 2/3*2(tan(t/2)
evaluating from 0 to?/2
=1/3 (1)(2) + 4/3 = 2/3 + 4/3 = 2

Q: What am I doing wrong?
This is bothering me. I'd like to learn to use the tables properly when u is a fraction or multiple of the variable. I'd really appreciate your advice.
Click to expand...
 
\(\displaystyle \int \ sec^{n}(u)du \ = \ \frac{sec^{n-2}(u)tan(u)}{n-1} \ + \ \frac{n-2}{n-1}\int \ sec^{n-2}(u)du \ + \ C\)

\(\displaystyle Let \ u \ = \ \frac{t}{2}, \ du \ = \ \frac{dt}{2}, \ 2du \ = \ dt.\)

\(\displaystyle Hence \ \int_{0}^{\frac{\pi}{2}} \ sec^{4}(\frac{t}{2})dt \ = \ 2\int_ {0}^{\frac{\pi}{4}}sec^{4}(u)du\)

\(\displaystyle Ergo, \ 2\int_ {0}^{\frac{\pi}{4}}sec^{4}(u)du \ = \ 2[\frac{sec^{2}(u)tan(u)}{3}|_{0}^{\frac{\pi}{4}} \ + \ \frac{2}{3}\int_{0}^{\frac{\pi}{4}}sec^{2}(u)du}]\)

\(\displaystyle 2[\frac{sec^{2}(u)tan(u)}{3} \ + \ \frac{2}{3}tan(u)}]_{0}^{\frac{\pi}{4}} \ = \ 2[\frac{(2)(1)}{3} \ + \ \frac{(2)(1)}{3}] = \ 2(\frac{4}{3}}) \ = \ \frac{8}{3} \ . \ QED\)
 
Thank you so much!

Now I understand where I was going wrong. That really makes things a lot clearer.

Thanks! I’ll keep practicing!
 
\(\displaystyle ..... \ QED\) .... Ergo .... QED ... Going Latin on us!!! :mrgreen: [/quote]
 
Subhotosh Khan, in order to break up any monotony ( whether real or imagined), ergo, one must resort to a poetic license.

Actually, you could say that even through English is considered a Teutonic language, our mother tongue has a lot of Latin and Greek in it.

Note: Ergo is Latin, I somehow thought it was a corruption of ere I go, ere meaning before, as for example how many times (if ever) have you heard someone say "It's four of the clock"which has been contracted to "It's four o' clock".
 
And as for you HappyCalculusStudent, glad I could be of assistance, as the last thing I want is for you to become an UnhappyCalculusStudent.
 
You must log in or register to reply here.
Top