Question

Integration
Help please

Answer 1

\[\int\limits dx/4\cos^32x-3\cos2x\]

Answer 2

the answer is :\[1/6 \ \\log[\sec6x+\tan6x]+C\]

Answer 3

how will i get this answer
please help

Answer 4

\[\int\limits\limits \frac{ 1 }{ 4\cos^3(2x)-3\cos(2x) }dx?\]

Answer 5

yes

Answer 6

look up formula of cos(3x)

Answer 7

4cos^x-3cosx right?

Answer 8

yeah cos^3x :)

Answer 9

yeah :)
sorry

Answer 10

So then you can just do a u - sub and then let cos(3u) = 3cos^3(u)-3cos(u), rational you genius!

Answer 11

4cos*

Answer 12

so
what will it be like?

Answer 13

u = 2x

Answer 14

then you can integrate 1/cos(3u) normaly

Answer 15

\[\frac{ 1 }{ 2 } \int\limits \frac{ 1 }{ \cos(3u) }du\]

Answer 16

hey
its cos6x

Answer 17

Just work it out and I'm sure you'll get it eventually :)

Answer 18

i mean