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Integrate tan3x

i cant do it

integrate tan3x please

P3 exam on monday!! AH!!!

The Integral of Tan(x) is -ln(Cos(x))

Because Tan(x) = Sin(x)/Cos(x) = -f'(x)/f(x) where f(x) = Cos(x)

Then you simply need to use substitution of 3x for t, or do it directly to obtain

Int[tan(3x)] = -1/3ln(Cos(x))

Reply 2

C4>O7

Not important but I think you normally see the integral of tanx written as 'ln|secx|'

Reply 3

jpowell

Yep also C4 raises a point I missed, you need to use modulus signs instead of the brackets I used.

Reply 4

ClaireC

i dont understand the -f'(x)/f(x) bit

could you do it out step by step?

and do the t thing too

sorry, im thick lol

Reply 5

Gaz031

ClaireC

i cant do it

integrate tan3x please

P3 exam on monday!! AH!!!

INT tan3x dx = INT sin3x/cos3x dx
u=cos3x. du/dx=-3sin3x. dx=du/(-3sin3x)
INT sin3x/cos3x dx = INT (sin3x)/u . du/(-3sin3x)
= (-1/3)INT 1/u du
= (-1/3)lnu + C
= (-1/3)lncos3x+C
= (1/3)ln(1/cos3x)+C
= (1/3)ln(sec3x)+C

Good luck with the P3 exam.

Reply 6

ClaireC

excellent gaz, i totally understand that. Thanks so much!!!

Reply 7

jmzcherry

tan3x is the same as sin3x/cos3x. sinx is the first dirivitive (excuse spellings) of cosx. i.e y=cosx, dy/dx=sinx.

Let cos3x=t

dt/dx=3sin3x so dt=3sin3x dx

1/3dt=sin3x

therefore the integral of sin3x/cos3x dx is: 1/3 X 1/t dt or 1/3 X t^-1 dt

1/3 X t-1 integrates to 1/3 X -lnt

the minus at the front of the ln can be moved to give 1/3lnt^-1 which is lncos^-1 x or ln(1/cos3x)

1/cos3x is sec3x hence ln|sec3x|

Correct me if im wrong but i believe than it is ln|sec3x| and not ln|secx|