Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

what is the next step? could somebody help me?

Answer 5

Use the reduction formula for m=4
\[\int\limits \sec^m(x)dx=\frac{\sin(x)\sec^m-1(x)}{m-1}+\frac{m-2}{m-1}\int\limits \sec^{m-2}(x)dx\]

Answer 6

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

Answer 7

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

Answer 8

now let m=4

Answer 9

\[\frac{1}{2}\int\limits\limits\limits\limits \sec^4(u)du=\frac{\sin(u)\sec^{4-1}(u)}{4-1}+\frac{4-2}{4-1}\int\limits\limits\limits\limits \sec^{4-2}(u)du\]

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

u =2x+1

Answer 14

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

Answer 15

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...

Answer 16

An alternative way to solve this problem

Answer 17

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

Answer 18

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

Answer 19

expand the function and distribute the integral

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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...

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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)

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

now can u see the full picture...

Answer 32

(v^3)/3+c

Answer 33

but v=tan(u)

Answer 34

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

Answer 35

hope this clarifies the point...

Answer 36

are u there?

Answer 37

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?

Answer 38

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...

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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 ..?

Answer 45

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
\[f(x) \neq f(-x)\]

Answer 46

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...

Answer 47

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...

Answer 48

: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?

Answer 49

nope they are different and depends, it could be confusing if u don't have experience, i would say try solving the elementary way...

Answer 50

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??

Answer 51

that was the integration

Answer 52

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

Answer 53

now when u differentiate tan u get sec

Answer 54

let me show u

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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??

Answer 62

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

Answer 63

is that what u meant? or i misunderstood?

Answer 64

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

Answer 65

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

Answer 66

tan^4(u)/4 +c

Answer 67

v^99

Answer 68

tan^100(u)/100+c

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

is that correct??

Answer 74

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

Answer 75

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

Answer 76

what is the integral of sin(x)

Answer 77

its: -cos (u) +c :D

Answer 78

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

Answer 79

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

Answer 80

no never do that

Answer 81

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

Answer 82

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

Answer 83

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

Answer 84

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

Answer 85

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

Answer 86

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

Answer 87

use the u sub and check for urself

Answer 88

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