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

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

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

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

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

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

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

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

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

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1now 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...

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

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

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

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

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

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

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1if 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
 one year 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.

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

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1when 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)

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

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

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

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

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

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

appleduardo
 one year 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?

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1Your 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...

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

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

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

appleduardo
 one year 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

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

appleduardo
 one year 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 ..?

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1An 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...

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1one 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
 one year 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?

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

appleduardo
 one year 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??

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

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

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

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

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

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

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

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

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

appleduardo
 one year 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??

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

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

appleduardo
 one year 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?

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

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

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

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

appleduardo
 one year 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 :\[\cos(x)+c\]

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

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

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

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

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

appleduardo
 one year 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:\[\frac{ 1 }{ 2 }\cos^2(x)+c\]

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

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

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

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

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

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

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

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