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

- appleduardo

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- appleduardo

\[\int\limits_{}^{}\sec^4(2x+1)dx\]

- appleduardo

so far ive done this--> \[\int\limits_{}^{}[1+\tan^2(2x)] [\sec^2(2x+1)]\]

- appleduardo

so then I get:\[\int\limits_{}^{}\sec^2(2x+1) +\int\limits_{}^{}\tan^2(2x+1)\sec^2(2x+1)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- appleduardo

what is the next step? could somebody help me?

- 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

now whatever answer u get multiply by 1/2 because
\[u =2x+1 \Longrightarrow du=2dx \longrightarrow dx=\frac{1}{2}du\]

- 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

now let m=4

- 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

\[=\frac{1}{2}[\frac{\sin(u)\sec^{3}(u)}{3}+\frac{2}{3}\int\limits\limits\limits\limits \sec^{2}(u)du]\]

- anonymous

\[=\frac{1}{6} \tan(u)\sec^2(x)+\frac{1}{3}\int\limits\limits \sec^2(u)du\]

- anonymous

\[=\frac{1}{6} \tan(u)\sec^2(u)+\frac{1}{3}\tan^2(u)+c\]

- anonymous

u =2x+1

- anonymous

\[=\frac{1}{6} \tan(2x+1)\sec^2(2x+1)+\frac{1}{3}\tan^2(2x+1)+c\]

- 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

An alternative way to solve this problem

- anonymous

there are many trig identities that can help in solving this problem one way is:

- anonymous

\[\int\limits \sec^2(u)\sec^2(u)du=\frac{1}{2}\int\limits \sec^2(u)(1+\tan^2(u))du\]

- anonymous

expand the function and distribute the integral

- anonymous

\[=\frac{1}{2}\int\limits \sec^2(u)du + \frac{1}{2} \int\limits \sec^2(u)\tan^2(u)du\]

- anonymous

\[=\frac{1}{2}\tan(u) +\frac{1}{6} \tan^3(u)+c\]

- anonymous

\[=\frac{1}{2}\tan(2x+1) +\frac{1}{6} \tan^3(2x+1)+c\]

- 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

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

ur welcome! i was expecting ur question i did that on purpose...

- anonymous

now there are two ways to think about this...

- 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

the other way to visualize this concept is by assuming that tan(u) =v

- anonymous

dv = sec^2(u)du -> du=dv/sec^(u)

- anonymous

\[\int\limits \tan^2(u)\sec^2(u)du=\int\limits v^2\sec^2(u)\frac{dv}{\sec^2(u)}\]

- anonymous

now can u see the full picture...

- anonymous

(v^3)/3+c

- anonymous

but v=tan(u)

- anonymous

=(1/3)tan^3(u)+c, where u = 2x+1

- anonymous

hope this clarifies the point...

- anonymous

are u there?

- 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

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

So u can correct the information for ur teacher, if u think this is what she/he meant...

- anonymous

but anyways i mean if u r not familiar with the even and odd functions then use the 2nd method...

- anonymous

by the way this problem can be solved in many ways...

- appleduardo

uhmm yeah i think i have to understand the second method first. i'll have to read it once again

- anonymous

just write it down on a piece of paper, you'll see how easy it is...

- 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

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

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

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

: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

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

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

that was the integration

- anonymous

the integral of tan^2(u)sec^(u)du

- anonymous

now when u differentiate tan u get sec

- anonymous

let me show u

- anonymous

\[\frac{1}{2} \int\limits\limits \sec^2(u)\tan^2(u)du\]

- anonymous

let
\[v = \tan(u) \therefore dv = \sec^2(u)du, so \space du=\frac{dv}{\sec^2(u)}\]

- anonymous

\[= \frac{1}{2} \int\limits\limits \sec^2(u)v^2\frac{dv}{\sec^2(u)}\]

- anonymous

can u see how sec^2(u) is cancelled so u are left with

- anonymous

\[\frac{1}{2} \int\limits v^2dv\]

- anonymous

and u know the rest, or do u want me to finish it to the end...

- 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

yes tan(u)=v therefore tan^2(u)=v^2

- anonymous

is that what u meant? or i misunderstood?

- appleduardo

yeep! :D thanks, so if tan^2 were tan^3 then "V" would equal to v^3 in the final result?

- anonymous

well in the case of the question above v^4/4+c

- anonymous

tan^4(u)/4 +c

- anonymous

v^99

- anonymous

tan^100(u)/100+c

- anonymous

if u did not have the sec^2(u), then you can't do that...

- anonymous

that's what i meant of the function multiplied by its derivative

- anonymous

if i tell u to integrate sin(x)cos(x), what answer would i get?

- appleduardo

i think i'd use -->:\[\int\limits_{}^{}sen udu\] and since u=sen, and du=cos, then :\[-\cos(x)+c\]

- appleduardo

is that correct??

- anonymous

u mean -(1/2)cos^2(x)+c

- anonymous

this is the whole point always when u integrate check if the function has its derivative...

- anonymous

what is the integral of sin(x)

- appleduardo

its: -cos (u) +c :D

- anonymous

and what result did u get for sin(x)cos(x) again?

- appleduardo

uhmm i thought the result was âˆ’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

no never do that

- anonymous

a rcipical of cos never gives u a minus sign be careful never make such a mistake !

- anonymous

\[\frac{1}{2}\frac{1}{\cos^2(x)}=\frac{1}{2}(\cos(x))^{-2}\]

- appleduardo

:o oohh, so is that the final result? +c

- anonymous

no that is the result of what you thought it was the result which is not the result in this case

- anonymous

\[\int\limits \sin(x)\cos(x)dx = -\frac{1}{2}\cos^2(x)+c\]

- appleduardo

but how did u get the 1/2 and cos^2 ?

- anonymous

use the u sub and check for urself

- appleduardo

yeep thanks! i'll try it :D :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.