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what is the integral of the following function? > sec^4 (2x+1)dx I hope somebody can help me out.
 one year ago
 one year ago
what is the integral of the following function? > sec^4 (2x+1)dx I hope somebody can help me out.
 one year ago
 one year ago

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appleduardoBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\sec^4(2x+1)dx\]
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
so far ive done this> \[\int\limits_{}^{}[1+\tan^2(2x)] [\sec^2(2x+1)]\]
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
so then I get:\[\int\limits_{}^{}\sec^2(2x+1) +\int\limits_{}^{}\tan^2(2x+1)\sec^2(2x+1)\]
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
what is the next step? could somebody help me?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
Use the reduction formula for m=4 \[\int\limits \sec^m(x)dx=\frac{\sin(x)\sec^m1(x)}{m1}+\frac{m2}{m1}\int\limits \sec^{m2}(x)dx\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
now whatever answer u get multiply by 1/2 because \[u =2x+1 \Longrightarrow du=2dx \longrightarrow dx=\frac{1}{2}du\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[\frac{1}{2}\int\limits\limits \sec^m(u)du=\frac{\sin(u)\sec^{m1}(u)}{m1}+\frac{m2}{m1}\int\limits\limits \sec^{m2}(u)du\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[\frac{1}{2}\int\limits\limits\limits\limits \sec^4(u)du=\frac{\sin(u)\sec^{41}(u)}{41}+\frac{42}{41}\int\limits\limits\limits\limits \sec^{42}(u)du\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[=\frac{1}{2}[\frac{\sin(u)\sec^{3}(u)}{3}+\frac{2}{3}\int\limits\limits\limits\limits \sec^{2}(u)du]\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[=\frac{1}{6} \tan(u)\sec^2(x)+\frac{1}{3}\int\limits\limits \sec^2(u)du\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[=\frac{1}{6} \tan(u)\sec^2(u)+\frac{1}{3}\tan^2(u)+c\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[=\frac{1}{6} \tan(2x+1)\sec^2(2x+1)+\frac{1}{3}\tan^2(2x+1)+c\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
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...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
An alternative way to solve this problem
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
there are many trig identities that can help in solving this problem one way is:
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[\int\limits \sec^2(u)\sec^2(u)du=\frac{1}{2}\int\limits \sec^2(u)(1+\tan^2(u))du\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
expand the function and distribute the integral
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[=\frac{1}{2}\int\limits \sec^2(u)du + \frac{1}{2} \int\limits \sec^2(u)\tan^2(u)du\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[=\frac{1}{2}\tan(u) +\frac{1}{6} \tan^3(u)+c\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[=\frac{1}{2}\tan(2x+1) +\frac{1}{6} \tan^3(2x+1)+c\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
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...
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
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.
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
ur welcome! i was expecting ur question i did that on purpose...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
now there are two ways to think about this...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
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)
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
the other way to visualize this concept is by assuming that tan(u) =v
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
dv = sec^2(u)du > du=dv/sec^(u)
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[\int\limits \tan^2(u)\sec^2(u)du=\int\limits v^2\sec^2(u)\frac{dv}{\sec^2(u)}\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
now can u see the full picture...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
=(1/3)tan^3(u)+c, where u = 2x+1
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
hope this clarifies the point...
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
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?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
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...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
So u can correct the information for ur teacher, if u think this is what she/he meant...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
but anyways i mean if u r not familiar with the even and odd functions then use the 2nd method...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
by the way this problem can be solved in many ways...
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
uhmm yeah i think i have to understand the second method first. i'll have to read it once again
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
just write it down on a piece of paper, you'll see how easy it is...
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
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 ..?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
if f(x) = (fx) 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 \[f(x) \neq f(x)\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
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 year ago

mathsmindBest ResponseYou've already chosen the best response.1
one of the easiest to test if the function is odd or even is visualizing the graph, if the graph is symmetrical about the yaxis 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...
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
: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?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
nope they are different and depends, it could be confusing if u don't have experience, i would say try solving the elementary way...
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
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??
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
that was the integration
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
the integral of tan^2(u)sec^(u)du
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
now when u differentiate tan u get sec
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[\frac{1}{2} \int\limits\limits \sec^2(u)\tan^2(u)du\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
let \[v = \tan(u) \therefore dv = \sec^2(u)du, so \space du=\frac{dv}{\sec^2(u)}\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[= \frac{1}{2} \int\limits\limits \sec^2(u)v^2\frac{dv}{\sec^2(u)}\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
can u see how sec^2(u) is cancelled so u are left with
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[\frac{1}{2} \int\limits v^2dv\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
and u know the rest, or do u want me to finish it to the end...
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
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??
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
yes tan(u)=v therefore tan^2(u)=v^2
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
is that what u meant? or i misunderstood?
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
yeep! :D thanks, so if tan^2 were tan^3 then "V" would equal to v^3 in the final result?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
well in the case of the question above v^4/4+c
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
if u did not have the sec^2(u), then you can't do that...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
that's what i meant of the function multiplied by its derivative
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
if i tell u to integrate sin(x)cos(x), what answer would i get?
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
i think i'd use >:\[\int\limits_{}^{}sen udu\] and since u=sen, and du=cos, then :\[\cos(x)+c\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
u mean (1/2)cos^2(x)+c
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
this is the whole point always when u integrate check if the function has its derivative...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
what is the integral of sin(x)
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
its: cos (u) +c :D
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
and what result did u get for sin(x)cos(x) again?
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
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:\[\frac{ 1 }{ 2 }\cos^2(x)+c\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
a rcipical of cos never gives u a minus sign be careful never make such a mistake !
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[\frac{1}{2}\frac{1}{\cos^2(x)}=\frac{1}{2}(\cos(x))^{2}\]
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
:o oohh, so is that the final result? +c
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
no that is the result of what you thought it was the result which is not the result in this case
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
\[\int\limits \sin(x)\cos(x)dx = \frac{1}{2}\cos^2(x)+c\]
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
but how did u get the 1/2 and cos^2 ?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.1
use the u sub and check for urself
 one year ago

appleduardoBest ResponseYou've already chosen the best response.0
yeep thanks! i'll try it :D :D
 one year ago
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