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appleduardo
what is the integral of the following function? --> sec^4 (2x+1)dx I hope somebody can help me out.
\[\int\limits_{}^{}\sec^4(2x+1)dx\]
so far ive done this--> \[\int\limits_{}^{}[1+\tan^2(2x)] [\sec^2(2x+1)]\]
so then I get:\[\int\limits_{}^{}\sec^2(2x+1) +\int\limits_{}^{}\tan^2(2x+1)\sec^2(2x+1)\]
what is the next step? could somebody help me?
Use the reduction formula for m=4 \[\int\limits \sec^m(x)dx=\frac{\sin(x)\sec^m-1(x)}{m-1}+\frac{m-2}{m-1}\int\limits \sec^{m-2}(x)dx\]
now whatever answer u get multiply by 1/2 because \[u =2x+1 \Longrightarrow du=2dx \longrightarrow dx=\frac{1}{2}du\]
\[\frac{1}{2}\int\limits\limits \sec^m(u)du=\frac{\sin(u)\sec^{m-1}(u)}{m-1}+\frac{m-2}{m-1}\int\limits\limits \sec^{m-2}(u)du\]
\[\frac{1}{2}\int\limits\limits\limits\limits \sec^4(u)du=\frac{\sin(u)\sec^{4-1}(u)}{4-1}+\frac{4-2}{4-1}\int\limits\limits\limits\limits \sec^{4-2}(u)du\]
\[=\frac{1}{2}[\frac{\sin(u)\sec^{3}(u)}{3}+\frac{2}{3}\int\limits\limits\limits\limits \sec^{2}(u)du]\]
\[=\frac{1}{6} \tan(u)\sec^2(x)+\frac{1}{3}\int\limits\limits \sec^2(u)du\]
\[=\frac{1}{6} \tan(u)\sec^2(u)+\frac{1}{3}\tan^2(u)+c\]
\[=\frac{1}{6} \tan(2x+1)\sec^2(2x+1)+\frac{1}{3}\tan^2(2x+1)+c\]
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...
An alternative way to solve this problem
there are many trig identities that can help in solving this problem one way is:
\[\int\limits \sec^2(u)\sec^2(u)du=\frac{1}{2}\int\limits \sec^2(u)(1+\tan^2(u))du\]
expand the function and distribute the integral
\[=\frac{1}{2}\int\limits \sec^2(u)du + \frac{1}{2} \int\limits \sec^2(u)\tan^2(u)du\]
\[=\frac{1}{2}\tan(u) +\frac{1}{6} \tan^3(u)+c\]
\[=\frac{1}{2}\tan(2x+1) +\frac{1}{6} \tan^3(2x+1)+c\]
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...
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.
ur welcome! i was expecting ur question i did that on purpose...
now there are two ways to think about this...
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)
the other way to visualize this concept is by assuming that tan(u) =v
dv = sec^2(u)du -> du=dv/sec^(u)
\[\int\limits \tan^2(u)\sec^2(u)du=\int\limits v^2\sec^2(u)\frac{dv}{\sec^2(u)}\]
now can u see the full picture...
=(1/3)tan^3(u)+c, where u = 2x+1
hope this clarifies the point...
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?
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...
So u can correct the information for ur teacher, if u think this is what she/he meant...
but anyways i mean if u r not familiar with the even and odd functions then use the 2nd method...
by the way this problem can be solved in many ways...
uhmm yeah i think i have to understand the second method first. i'll have to read it once again
just write it down on a piece of paper, you'll see how easy it is...
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 ..?
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 \[f(x) \neq f(-x)\]
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...
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...
: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?
nope they are different and depends, it could be confusing if u don't have experience, i would say try solving the elementary way...
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??
that was the integration
the integral of tan^2(u)sec^(u)du
now when u differentiate tan u get sec
\[\frac{1}{2} \int\limits\limits \sec^2(u)\tan^2(u)du\]
let \[v = \tan(u) \therefore dv = \sec^2(u)du, so \space du=\frac{dv}{\sec^2(u)}\]
\[= \frac{1}{2} \int\limits\limits \sec^2(u)v^2\frac{dv}{\sec^2(u)}\]
can u see how sec^2(u) is cancelled so u are left with
\[\frac{1}{2} \int\limits v^2dv\]
and u know the rest, or do u want me to finish it to the end...
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??
yes tan(u)=v therefore tan^2(u)=v^2
is that what u meant? or i misunderstood?
yeep! :D thanks, so if tan^2 were tan^3 then "V" would equal to v^3 in the final result?
well in the case of the question above v^4/4+c
if u did not have the sec^2(u), then you can't do that...
that's what i meant of the function multiplied by its derivative
if i tell u to integrate sin(x)cos(x), what answer would i get?
i think i'd use -->:\[\int\limits_{}^{}sen udu\] and since u=sen, and du=cos, then :\[-\cos(x)+c\]
u mean -(1/2)cos^2(x)+c
this is the whole point always when u integrate check if the function has its derivative...
what is the integral of sin(x)
its: -cos (u) +c :D
and what result did u get for sin(x)cos(x) again?
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:\[-\frac{ 1 }{ 2 }\cos^2(x)+c\]
a rcipical of cos never gives u a minus sign be careful never make such a mistake !
\[\frac{1}{2}\frac{1}{\cos^2(x)}=\frac{1}{2}(\cos(x))^{-2}\]
:o oohh, so is that the final result? +c
no that is the result of what you thought it was the result which is not the result in this case
\[\int\limits \sin(x)\cos(x)dx = -\frac{1}{2}\cos^2(x)+c\]
but how did u get the 1/2 and cos^2 ?
use the u sub and check for urself
yeep thanks! i'll try it :D :D