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

1. appleduardo

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

2. appleduardo

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

3. appleduardo

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

4. appleduardo

what is the next step? could somebody help me?

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

now let m=4

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

u =2x+1

14. anonymous

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

15. 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...

16. anonymous

An alternative way to solve this problem

17. anonymous

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

18. anonymous

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

19. anonymous

expand the function and distribute the integral

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. 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...

24. 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.

25. anonymous

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

26. anonymous

27. 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)

28. anonymous

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

29. anonymous

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

30. anonymous

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

31. anonymous

now can u see the full picture...

32. anonymous

(v^3)/3+c

33. anonymous

but v=tan(u)

34. anonymous

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

35. anonymous

hope this clarifies the point...

36. anonymous

are u there?

37. 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?

38. 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...

39. anonymous

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

40. anonymous

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

41. anonymous

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

42. appleduardo

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

43. anonymous

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

44. 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 ..?

45. 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 $f(x) \neq f(-x)$

46. 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...

47. 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...

48. 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?

49. 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...

50. 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??

51. anonymous

that was the integration

52. anonymous

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

53. anonymous

now when u differentiate tan u get sec

54. anonymous

let me show u

55. anonymous

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

56. anonymous

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

57. anonymous

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

58. anonymous

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

59. anonymous

$\frac{1}{2} \int\limits v^2dv$

60. anonymous

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

61. 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??

62. anonymous

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

63. anonymous

is that what u meant? or i misunderstood?

64. appleduardo

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

65. anonymous

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

66. anonymous

tan^4(u)/4 +c

67. anonymous

v^99

68. anonymous

tan^100(u)/100+c

69. anonymous

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

70. anonymous

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

71. anonymous

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

72. appleduardo

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

73. appleduardo

is that correct??

74. anonymous

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

75. anonymous

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

76. anonymous

what is the integral of sin(x)

77. appleduardo

its: -cos (u) +c :D

78. anonymous

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

79. appleduardo

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

80. anonymous

no never do that

81. anonymous

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

82. anonymous

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

83. appleduardo

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

84. anonymous

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

85. anonymous

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

86. appleduardo

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

87. anonymous

use the u sub and check for urself

88. appleduardo

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