• Support PF! Buy your school textbooks, materials and every day products Here!

Is my book wrong?

  • Thread starter mad
  • Start date
  • #1
mad
65
0
I have this integration problem that I did but it doesnt give me the right answer. But there are like 3 other similar exercices I did the same way and I got all the right answers.. maybe it is my book (I dont think so ... :p)

It's an integration problem:

[tex]\int tg^3(4x) sec^4(4x)dx[/tex]

heres what I did:

[tex]\int (sec^2(4x)-1)(tg(4x))(sec^4(4x))dx[/tex]
= [tex]\int (sec^6(4x) - sec^4(4x)) (tg4x) dx [/tex]

u= sec 4x
du = 4(sec4x)(tg4x)dx --> dx = du/(4(sec4x)(tg4x))

=[tex]\int \frac{(u^6 - u^4) du}{4u} [/tex]

(replaced the sec(4x) at denom. with u since u=sec4x)

=[tex]\frac{1}{4}\int u^5 - [/tex] [tex]\frac{1}{4}\int u^3 [/tex]
= [tex]\frac{1}{24}sec^6(4x) -[/tex] [tex]\frac{1}{16} sec^4(4x) [/tex] +C


Add anything you want! Thanks everyone

BTW the answer in my book is
(1/16) tg^4 (4x) + (1/24) tg^6(4x)
I tried it in my calc with an x and it doesnt give the same answer.
 
Last edited:

Answers and Replies

  • #2
Imo
30
0
Actually, those aer the same answers (up to that constant, if you put it directly into your calculator, they differ already by a constant that doesn't matter in this type of integration...add 1/48 to your answer to get the other). They just went another direction in the integration.
 
  • #3
2,209
1
What the hell is tg(x)?
 
  • #4
dextercioby
Science Advisor
Homework Helper
Insights Author
12,977
540
Here's another way.Denote the integral by I and write everything in terms of sine a cosine...

[tex] I=\int \left(\frac{\sin^{3}4x}{\cos^{7}4x}\right) \ dx [/tex]

[tex] 4x=u \Rightarrow 4dx=du [/tex]

[tex] I=\frac{1}{4}\int \left(\frac{\sin^{4}u}{\cos^{7}u}\right) \ du
=-\frac{1}{4}\int \left(\frac{1-\cos^{2}u}{\cos^{7}u}\right) \ d(\cos u) =-\frac{1}{4}\left[\frac{(\cos u)^{-6}}{-6}-\frac{(\cos u)^{-4}}{-4}\right] +\mathcal{C}[/tex]

Therefore,reversing the substitution made

[tex]I=\frac{1}{24}\frac{1}{\cos^{6}4x}-\frac{1}{16}\frac{1}{\cos^{4}4x}+\mathcal{C} [/tex]

Daniel.
 
  • #5
dextercioby
Science Advisor
Homework Helper
Insights Author
12,977
540
whozum said:
What the hell is tg(x)?
Tangent of (x),what else??? :rolleyes:

Daniel.
 
  • #6
mad
65
0
dextercioby said:
Here's another way.Denote the integral by I and write everything in terms of sine a cosine...

[tex] I=\int \left(\frac{\sin^{3}4x}{\cos^{7}4x}\right) \ dx [/tex]

[tex] 4x=u \Rightarrow 4dx=du [/tex]

[tex] I=\frac{1}{4}\int \left(\frac{\sin^{4}u}{\cos^{7}u}\right) \ du
=-\frac{1}{4}\int \left(\frac{1-\cos^{2}u}{\cos^{7}u}\right) \ d(\cos u) =-\frac{1}{4}\left[\frac{(\cos u)^{-6}}{-6}-\frac{(\cos u)^{-4}}{-4}\right] +\mathcal{C}[/tex]

Therefore,reversing the substitution made

[tex]I=\frac{1}{24}\frac{1}{\cos^{6}4x}-\frac{1}{16}\frac{1}{\cos^{4}4x}+\mathcal{C} [/tex]

Daniel.

Thanks for your help, Daniel.
I see my solution was okay. We had to do the problem by exponents of sec and tg. (sorry I dont know what the method is called in english)

However, what does d(cos u) means in your solution? and I know you replaced a sin^4 (u) by (1-cos^2 (u)), but where is the other sin^2 (x)
Surely it is the d(cos u) you used, but I'm not familiar with this notation
 
Last edited:
  • #7
2,209
1
dextercioby said:
Tangent of (x),what else??? :rolleyes:

Daniel.
What happened to tan(x)?
 
  • #8
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
35
whozum said:
What happened to tan(x)?
Skin cancer you know...the poor devil.
 

Related Threads for: Is my book wrong?

  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
15
Views
2K
Replies
12
Views
2K
Replies
3
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
Top