appleduardo
  • appleduardo
what is the integral of the following function? --> sec^4 (2x+1)dx I hope somebody can help me out.
Mathematics
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schrodinger
  • schrodinger
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appleduardo
  • appleduardo
\[\int\limits_{}^{}\sec^4(2x+1)dx\]
appleduardo
  • appleduardo
so far ive done this--> \[\int\limits_{}^{}[1+\tan^2(2x)] [\sec^2(2x+1)]\]
appleduardo
  • appleduardo
so then I get:\[\int\limits_{}^{}\sec^2(2x+1) +\int\limits_{}^{}\tan^2(2x+1)\sec^2(2x+1)\]

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appleduardo
  • appleduardo
what is the next step? could somebody help me?
anonymous
  • anonymous
Use the reduction formula for m=4 \[\int\limits \sec^m(x)dx=\frac{\sin(x)\sec^m-1(x)}{m-1}+\frac{m-2}{m-1}\int\limits \sec^{m-2}(x)dx\]
anonymous
  • anonymous
now whatever answer u get multiply by 1/2 because \[u =2x+1 \Longrightarrow du=2dx \longrightarrow dx=\frac{1}{2}du\]
anonymous
  • anonymous
\[\frac{1}{2}\int\limits\limits \sec^m(u)du=\frac{\sin(u)\sec^{m-1}(u)}{m-1}+\frac{m-2}{m-1}\int\limits\limits \sec^{m-2}(u)du\]
anonymous
  • anonymous
now let m=4
anonymous
  • anonymous
\[\frac{1}{2}\int\limits\limits\limits\limits \sec^4(u)du=\frac{\sin(u)\sec^{4-1}(u)}{4-1}+\frac{4-2}{4-1}\int\limits\limits\limits\limits \sec^{4-2}(u)du\]
anonymous
  • anonymous
\[=\frac{1}{2}[\frac{\sin(u)\sec^{3}(u)}{3}+\frac{2}{3}\int\limits\limits\limits\limits \sec^{2}(u)du]\]
anonymous
  • anonymous
\[=\frac{1}{6} \tan(u)\sec^2(x)+\frac{1}{3}\int\limits\limits \sec^2(u)du\]
anonymous
  • anonymous
\[=\frac{1}{6} \tan(u)\sec^2(u)+\frac{1}{3}\tan^2(u)+c\]
anonymous
  • anonymous
u =2x+1
anonymous
  • anonymous
\[=\frac{1}{6} \tan(2x+1)\sec^2(2x+1)+\frac{1}{3}\tan^2(2x+1)+c\]
anonymous
  • anonymous
now using the reduction formula is useful because it reduces the amount of repetition of integration by parts, and simplifies the solution with less probability of errors...
anonymous
  • anonymous
An alternative way to solve this problem
anonymous
  • anonymous
there are many trig identities that can help in solving this problem one way is:
anonymous
  • anonymous
\[\int\limits \sec^2(u)\sec^2(u)du=\frac{1}{2}\int\limits \sec^2(u)(1+\tan^2(u))du\]
anonymous
  • anonymous
expand the function and distribute the integral
anonymous
  • anonymous
\[=\frac{1}{2}\int\limits \sec^2(u)du + \frac{1}{2} \int\limits \sec^2(u)\tan^2(u)du\]
anonymous
  • anonymous
\[=\frac{1}{2}\tan(u) +\frac{1}{6} \tan^3(u)+c\]
anonymous
  • anonymous
\[=\frac{1}{2}\tan(2x+1) +\frac{1}{6} \tan^3(2x+1)+c\]
anonymous
  • anonymous
if u think that the answers are different in the two methods, then that is not the case they are both the same u may use the trig identities to justify both solutions...
appleduardo
  • appleduardo
thank you very much mathsmind! i was kind of stressed, :/ but i have one queation:how did u get tan^3 ??in the final result.
anonymous
  • anonymous
ur welcome! i was expecting ur question i did that on purpose...
anonymous
  • anonymous
now there are two ways to think about this...
anonymous
  • anonymous
when u see a function with its derivative then u use the normal 1/(n+1tan^(n+1) like we do with 1/(n+1)x^(n+1)
anonymous
  • anonymous
the other way to visualize this concept is by assuming that tan(u) =v
anonymous
  • anonymous
dv = sec^2(u)du -> du=dv/sec^(u)
anonymous
  • anonymous
\[\int\limits \tan^2(u)\sec^2(u)du=\int\limits v^2\sec^2(u)\frac{dv}{\sec^2(u)}\]
anonymous
  • anonymous
now can u see the full picture...
anonymous
  • anonymous
(v^3)/3+c
anonymous
  • anonymous
but v=tan(u)
anonymous
  • anonymous
=(1/3)tan^3(u)+c, where u = 2x+1
anonymous
  • anonymous
hope this clarifies the point...
anonymous
  • anonymous
are u there?
appleduardo
  • appleduardo
yeep i am still here! i am trying to understand what you just typed! :P. uhmm i have another question, uhmm my teacher once told me that we can only use the reduction formula when "n" isnt a pair number, and in this case n=4, so uhmm is it right to solve it with that formula anyways?
anonymous
  • anonymous
Your teacher must have gone into a detailed answer, what happens is u need to consider if the function is an even one or an odd one, so because sec is an even function then u can use an even power...
anonymous
  • anonymous
So u can correct the information for ur teacher, if u think this is what she/he meant...
anonymous
  • anonymous
but anyways i mean if u r not familiar with the even and odd functions then use the 2nd method...
anonymous
  • anonymous
by the way this problem can be solved in many ways...
appleduardo
  • appleduardo
uhmm yeah i think i have to understand the second method first. i'll have to read it once again
anonymous
  • anonymous
just write it down on a piece of paper, you'll see how easy it is...
appleduardo
  • appleduardo
omG i just did and i understood it all! :D thank you very much mathsmind!, now id like to understand the reduction method, so uhmm what does "even" and "odd" fuctions mean ..?
anonymous
  • anonymous
if f(x) = (f-x) then that function is said to be even, e.g. x^2, x^4, x^6...etc or you may say x^n, where n = even number, also cos(x) is an even function, but u have to be careful u always need to perform a test and not make a mistake because for example (x+4)^2 is not an even function because \[f(x) \neq f(-x)\]
anonymous
  • anonymous
An odd function on the other hand is defined as -f(x)=f(-x), such as x^3, x^5, x^7, and sin(x) is also an odd function but x^3+1 is not an odd function...
anonymous
  • anonymous
one of the easiest to test if the function is odd or even is visualizing the graph, if the graph is symmetrical about the y-axis then its an even function, but if the graph is symmetrical about y=x then its an odd function, in other words symmetrical about the origin, however there are cases where the functions are neither odd or even...
appleduardo
  • appleduardo
:o cool!! :D but when i want to make use of the reduction formula.. is the same formula i have to use for an "odd" and an "even" function?
anonymous
  • anonymous
nope they are different and depends, it could be confusing if u don't have experience, i would say try solving the elementary way...
appleduardo
  • appleduardo
wow there's a lot of things to consider then and thank you so much for all your support. :P i have another question. When you solved this integral by the elementary way, you said tan^3 was the result of 1/(n+1)tan^(n+1) ... but why you did that? i mean, in order to get ...what??
anonymous
  • anonymous
that was the integration
anonymous
  • anonymous
the integral of tan^2(u)sec^(u)du
anonymous
  • anonymous
now when u differentiate tan u get sec
anonymous
  • anonymous
let me show u
anonymous
  • anonymous
\[\frac{1}{2} \int\limits\limits \sec^2(u)\tan^2(u)du\]
anonymous
  • anonymous
let \[v = \tan(u) \therefore dv = \sec^2(u)du, so \space du=\frac{dv}{\sec^2(u)}\]
anonymous
  • anonymous
\[= \frac{1}{2} \int\limits\limits \sec^2(u)v^2\frac{dv}{\sec^2(u)}\]
anonymous
  • anonymous
can u see how sec^2(u) is cancelled so u are left with
anonymous
  • anonymous
\[\frac{1}{2} \int\limits v^2dv\]
anonymous
  • anonymous
and u know the rest, or do u want me to finish it to the end...
appleduardo
  • appleduardo
yeep its alright! :D thank you! but if tan was originally tan^2 and then you took just tan to be your "u" function, it doesnt change the final result??
anonymous
  • anonymous
yes tan(u)=v therefore tan^2(u)=v^2
anonymous
  • anonymous
is that what u meant? or i misunderstood?
appleduardo
  • appleduardo
yeep! :D thanks, so if tan^2 were tan^3 then "V" would equal to v^3 in the final result?
anonymous
  • anonymous
well in the case of the question above v^4/4+c
anonymous
  • anonymous
tan^4(u)/4 +c
anonymous
  • anonymous
v^99
anonymous
  • anonymous
tan^100(u)/100+c
anonymous
  • anonymous
if u did not have the sec^2(u), then you can't do that...
anonymous
  • anonymous
that's what i meant of the function multiplied by its derivative
anonymous
  • anonymous
if i tell u to integrate sin(x)cos(x), what answer would i get?
appleduardo
  • appleduardo
i think i'd use -->:\[\int\limits_{}^{}sen udu\] and since u=sen, and du=cos, then :\[-\cos(x)+c\]
appleduardo
  • appleduardo
is that correct??
anonymous
  • anonymous
u mean -(1/2)cos^2(x)+c
anonymous
  • anonymous
this is the whole point always when u integrate check if the function has its derivative...
anonymous
  • anonymous
what is the integral of sin(x)
appleduardo
  • appleduardo
its: -cos (u) +c :D
anonymous
  • anonymous
and what result did u get for sin(x)cos(x) again?
appleduardo
  • appleduardo
uhmm i thought the result was −cos(x)+c , but if i make use of \[\frac{ 1 }{( n+1)(\cos ^{n+1})} =\frac{ 1 }{ 2\cos^2 }\] , then i'd get:\[-\frac{ 1 }{ 2 }\cos^2(x)+c\]
anonymous
  • anonymous
no never do that
anonymous
  • anonymous
a rcipical of cos never gives u a minus sign be careful never make such a mistake !
anonymous
  • anonymous
\[\frac{1}{2}\frac{1}{\cos^2(x)}=\frac{1}{2}(\cos(x))^{-2}\]
appleduardo
  • appleduardo
:o oohh, so is that the final result? +c
anonymous
  • anonymous
no that is the result of what you thought it was the result which is not the result in this case
anonymous
  • anonymous
\[\int\limits \sin(x)\cos(x)dx = -\frac{1}{2}\cos^2(x)+c\]
appleduardo
  • appleduardo
but how did u get the 1/2 and cos^2 ?
anonymous
  • anonymous
use the u sub and check for urself
appleduardo
  • appleduardo
yeep thanks! i'll try it :D :D

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