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

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

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

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

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0what is the next step? could somebody help me?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use the reduction formula for m=4 \[\int\limits \sec^m(x)dx=\frac{\sin(x)\sec^m1(x)}{m1}+\frac{m2}{m1}\int\limits \sec^{m2}(x)dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now whatever answer u get multiply by 1/2 because \[u =2x+1 \Longrightarrow du=2dx \longrightarrow dx=\frac{1}{2}du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\int\limits\limits \sec^m(u)du=\frac{\sin(u)\sec^{m1}(u)}{m1}+\frac{m2}{m1}\int\limits\limits \sec^{m2}(u)du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\int\limits\limits\limits\limits \sec^4(u)du=\frac{\sin(u)\sec^{41}(u)}{41}+\frac{42}{41}\int\limits\limits\limits\limits \sec^{42}(u)du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1}{2}[\frac{\sin(u)\sec^{3}(u)}{3}+\frac{2}{3}\int\limits\limits\limits\limits \sec^{2}(u)du]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1}{6} \tan(u)\sec^2(x)+\frac{1}{3}\int\limits\limits \sec^2(u)du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1}{6} \tan(u)\sec^2(u)+\frac{1}{3}\tan^2(u)+c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1}{6} \tan(2x+1)\sec^2(2x+1)+\frac{1}{3}\tan^2(2x+1)+c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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
 3 years ago
Best ResponseYou've already chosen the best response.0An alternative way to solve this problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there are many trig identities that can help in solving this problem one way is:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \sec^2(u)\sec^2(u)du=\frac{1}{2}\int\limits \sec^2(u)(1+\tan^2(u))du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0expand the function and distribute the integral

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1}{2}\int\limits \sec^2(u)du + \frac{1}{2} \int\limits \sec^2(u)\tan^2(u)du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1}{2}\tan(u) +\frac{1}{6} \tan^3(u)+c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1}{2}\tan(2x+1) +\frac{1}{6} \tan^3(2x+1)+c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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
 3 years ago
Best ResponseYou've already chosen the best response.0thank 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
 3 years ago
Best ResponseYou've already chosen the best response.0ur welcome! i was expecting ur question i did that on purpose...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now there are two ways to think about this...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when 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
 3 years ago
Best ResponseYou've already chosen the best response.0the other way to visualize this concept is by assuming that tan(u) =v

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dv = sec^2(u)du > du=dv/sec^(u)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \tan^2(u)\sec^2(u)du=\int\limits v^2\sec^2(u)\frac{dv}{\sec^2(u)}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now can u see the full picture...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0=(1/3)tan^3(u)+c, where u = 2x+1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hope this clarifies the point...

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0yeep 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
 3 years ago
Best ResponseYou've already chosen the best response.0Your 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
 3 years ago
Best ResponseYou've already chosen the best response.0So u can correct the information for ur teacher, if u think this is what she/he meant...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but anyways i mean if u r not familiar with the even and odd functions then use the 2nd method...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by the way this problem can be solved in many ways...

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0uhmm yeah i think i have to understand the second method first. i'll have to read it once again

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just write it down on a piece of paper, you'll see how easy it is...

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0omG 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
 3 years ago
Best ResponseYou've already chosen the best response.0if 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 \[f(x) \neq f(x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0An 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
 3 years ago
Best ResponseYou've already chosen the best response.0one 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...

appleduardo
 3 years ago
Best 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nope they are different and depends, it could be confusing if u don't have experience, i would say try solving the elementary way...

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0wow 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
 3 years ago
Best ResponseYou've already chosen the best response.0that was the integration

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the integral of tan^2(u)sec^(u)du

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now when u differentiate tan u get sec

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2} \int\limits\limits \sec^2(u)\tan^2(u)du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let \[v = \tan(u) \therefore dv = \sec^2(u)du, so \space du=\frac{dv}{\sec^2(u)}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[= \frac{1}{2} \int\limits\limits \sec^2(u)v^2\frac{dv}{\sec^2(u)}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u see how sec^2(u) is cancelled so u are left with

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2} \int\limits v^2dv\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and u know the rest, or do u want me to finish it to the end...

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0yeep 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
 3 years ago
Best ResponseYou've already chosen the best response.0yes tan(u)=v therefore tan^2(u)=v^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that what u meant? or i misunderstood?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well in the case of the question above v^4/4+c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if u did not have the sec^2(u), then you can't do that...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's what i meant of the function multiplied by its derivative

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if i tell u to integrate sin(x)cos(x), what answer would i get?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u mean (1/2)cos^2(x)+c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is the whole point always when u integrate check if the function has its derivative...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is the integral of sin(x)

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0its: cos (u) +c :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and what result did u get for sin(x)cos(x) again?

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0uhmm 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
 3 years ago
Best ResponseYou've already chosen the best response.0a rcipical of cos never gives u a minus sign be careful never make such a mistake !

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\frac{1}{\cos^2(x)}=\frac{1}{2}(\cos(x))^{2}\]

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0:o oohh, so is that the final result? +c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no that is the result of what you thought it was the result which is not the result in this case

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \sin(x)\cos(x)dx = \frac{1}{2}\cos^2(x)+c\]

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0but how did u get the 1/2 and cos^2 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use the u sub and check for urself

appleduardo
 3 years ago
Best ResponseYou've already chosen the best response.0yeep thanks! i'll try it :D :D
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