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

1. appleduardo Group Title

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

2. appleduardo Group Title

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

3. appleduardo Group Title

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

4. appleduardo Group Title

what is the next step? could somebody help me?

5. mathsmind Group Title

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. mathsmind Group Title

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

7. mathsmind Group Title

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

8. mathsmind Group Title

now let m=4

9. mathsmind Group Title

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

10. mathsmind Group Title

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

11. mathsmind Group Title

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

12. mathsmind Group Title

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

13. mathsmind Group Title

u =2x+1

14. mathsmind Group Title

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

15. mathsmind Group Title

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. mathsmind Group Title

An alternative way to solve this problem

17. mathsmind Group Title

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

18. mathsmind Group Title

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

19. mathsmind Group Title

expand the function and distribute the integral

20. mathsmind Group Title

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

21. mathsmind Group Title

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

22. mathsmind Group Title

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

23. mathsmind Group Title

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 Group Title

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. mathsmind Group Title

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

26. mathsmind Group Title

27. mathsmind Group Title

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. mathsmind Group Title

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

29. mathsmind Group Title

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

30. mathsmind Group Title

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

31. mathsmind Group Title

now can u see the full picture...

32. mathsmind Group Title

(v^3)/3+c

33. mathsmind Group Title

but v=tan(u)

34. mathsmind Group Title

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

35. mathsmind Group Title

hope this clarifies the point...

36. mathsmind Group Title

are u there?

37. appleduardo Group Title

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. mathsmind Group Title

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. mathsmind Group Title

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

40. mathsmind Group Title

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

41. mathsmind Group Title

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

42. appleduardo Group Title

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

43. mathsmind Group Title

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

44. appleduardo Group Title

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. mathsmind Group Title

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. mathsmind Group Title

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. mathsmind Group Title

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 Group Title

: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. mathsmind Group Title

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 Group Title

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. mathsmind Group Title

that was the integration

52. mathsmind Group Title

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

53. mathsmind Group Title

now when u differentiate tan u get sec

54. mathsmind Group Title

let me show u

55. mathsmind Group Title

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

56. mathsmind Group Title

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

57. mathsmind Group Title

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

58. mathsmind Group Title

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

59. mathsmind Group Title

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

60. mathsmind Group Title

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

61. appleduardo Group Title

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. mathsmind Group Title

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

63. mathsmind Group Title

is that what u meant? or i misunderstood?

64. appleduardo Group Title

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

65. mathsmind Group Title

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

66. mathsmind Group Title

tan^4(u)/4 +c

67. mathsmind Group Title

v^99

68. mathsmind Group Title

tan^100(u)/100+c

69. mathsmind Group Title

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

70. mathsmind Group Title

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

71. mathsmind Group Title

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

72. appleduardo Group Title

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

73. appleduardo Group Title

is that correct??

74. mathsmind Group Title

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

75. mathsmind Group Title

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

76. mathsmind Group Title

what is the integral of sin(x)

77. appleduardo Group Title

its: -cos (u) +c :D

78. mathsmind Group Title

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

79. appleduardo Group Title

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. mathsmind Group Title

no never do that

81. mathsmind Group Title

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

82. mathsmind Group Title

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

83. appleduardo Group Title

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

84. mathsmind Group Title

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

85. mathsmind Group Title

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

86. appleduardo Group Title

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

87. mathsmind Group Title

use the u sub and check for urself

88. appleduardo Group Title

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