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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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