- #1
mad
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I have this integration problem that I did but it doesn't 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 don't 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 doesn't give the same answer.
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 doesn't give the same answer.
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