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Integral of 4Tan3x, and AQA Formula booklet

Basically i was integrating Tan3x however in the AQA formula booklet it states that the integral of Tanx = ln|secx| which is fine but when i integrated Tan3x i made it equal to ln|sec3x| not 1/3ln|sec3x| which is what somebody said from a thread a while back.

Usually in the formula booklet it writes the integral of say sec^2kx as 1/k tankx, however it does not do this with the integral of Tanx which is why i did not stick a 1/3 at the front of my integral of Tan3x. I know i should divide by 3 when i integrate the function but i don't understand why the formula booklet has not indicated the 1/k part.

so basically is the Integral of 4Tan3x = 4(1/3)ln|sec3x| + c ? :smile:

I am sorry if this sounds confusing btw! :smile:
Reply 1
Avatar for raveen789
raveen789
OP
9
EDIT: ignore this sorry i found further down that the integral of tankx is something else, sorry! :smile:
Reply 2
Avatar for raheem94
raheem94
13
Original post by raveen789
Basically i was integrating Tan3x however in the AQA formula booklet it states that the integral of Tanx = ln|secx| which is fine but when i integrated Tan3x i made it equal to ln|sec3x| not 1/3ln|sec3x| which is what somebody said from a thread a while back.

Usually in the formula booklet it writes the integral of say sec^2kx as 1/k tankx, however it does not do this with the integral of Tanx which is why i did not stick a 1/3 at the front of my integral of Tan3x. I know i should divide by 3 when i integrate the function but i don't understand why the formula booklet has not indicated the 1/k part.

so basically is the Integral of 4Tan3x = 4(1/3)ln|sec3x| + c ? :smile:

I am sorry if this sounds confusing btw! :smile:


∫4tan3x dx=43ln∣sec3x∣+C \displaystyle \int 4tan3x \ dx = \frac43 ln|sec3x| + C

You are doing correct.
Reply 3
Avatar for raveen789
raveen789
OP
9
Original post by raheem94
∫4tan3x dx=43ln∣sec3x∣+C \displaystyle \int 4tan3x \ dx = \frac43 ln|sec3x| + C

You are doing correct.


oh it says that the integral of tanKx = ln|coskx| :/ which is quite confusing!
Reply 4
Avatar for raheem94
raheem94
13
Original post by raveen789
oh it says that the integral of tanKx = ln|coskx| :/ which is quite confusing!


It would probably be ∫tankx dx=−1kln∣coskx∣+C \displaystyle \int tankx \ dx = - \frac1k ln|coskx| + C
(edited 11 years ago)
Original post by raheem94
It would probably be ∫tankx=−1kln∣coskx∣+C \displaystyle \int tankx = - \frac1k ln|coskx| + C


Why the minus sign?
(edited 11 years ago)
Reply 6
Avatar for raheem94
raheem94
13
Original post by ghostwalker
Why the minus sign?


∫tankx dx=−1kln∣coskx∣+C \displaystyle \int tankx \ dx = - \frac1k ln|coskx| + C

−1kln∣coskx∣=1kln∣coskx∣−1=1kln∣seckx∣ \displaystyle - \frac1k ln|coskx| = \frac1k ln|coskx|^{-1} = \frac1k ln|seckx|

Please correct me if i am wrong.
(edited 11 years ago)
Original post by raheem94
∫tankx=−1kln∣coskx∣+C \displaystyle \int tankx = - \frac1k ln|coskx| + C

−1kln∣coskx∣=1kln∣coskx∣−1=1k∣seckx∣ \displaystyle - \frac1k ln|coskx| = \frac1k ln|coskx|^{-1} = \frac1k|seckx|

Please correct me if i am wrong.


Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).

Ignore: Irrelevant.
(edited 11 years ago)
Original post by ghostwalker
Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).


Maybe I'm missing something but I don't see the relevance? :s-smilie:
Reply 9
Avatar for raheem94
raheem94
13
Original post by ghostwalker
Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).


These aren't hyperbolic functions.

It is a 'k' not a 'h'.

Where is the typo, i don't see it?
Reply 10
Avatar for raheem94
raheem94
13
Original post by hassi94
Maybe I'm missing something but I don't see the relevance? :s-smilie:


Is the formula i have written correct?

Sorry for missing the 'dx', i forgot to write it.
Original post by raheem94
∫tankx=−1kln∣coskx∣+C \displaystyle \int tankx = - \frac1k ln|coskx| + C

−1kln∣coskx∣=1kln∣coskx∣−1=1k∣seckx∣ \displaystyle - \frac1k ln|coskx| = \frac1k ln|coskx|^{-1} = \frac1k|seckx|

Please correct me if i am wrong.


Original post by hassi94
Maybe I'm missing something but I don't see the relevance? :s-smilie:


I'm being a moron!

Reading k as h, and thinking it's hyperbolics. Duh!

Ignore me: I'll just go and shoot myself.
(edited 11 years ago)
Original post by raheem94
These aren't hyperbolic functions.

It is a 'k' not a 'h'.

Where is the typo, i don't see it?


The typo was you missing 'ln' on the last bit of the second line :smile:

Otherwise I think it's correct.
Original post by ghostwalker
I'm being a moron!

Reading k as h, and thinking it's hyperbolics. Duh!

Ignore me.


Aha okay that makes sense now, don't worry about it :tongue:
Original post by ghostwalker
I'm being a moron!



((((hugs)))) for ghostwalker

we have all been there
Reply 15
Avatar for raheem94
raheem94
13
Original post by ghostwalker
I'm being a moron!

Reading k as h, and thinking it's hyperbolics. Duh!

Ignore me.


Never knew you also make mistakes!

Original post by hassi94
The typo was you missing 'ln' on the last bit of the second line :smile:

Otherwise I think it's correct.


Thanks i will edit my post, its too difficult to spot your own mistakes.
Reply 16
Avatar for raveen789
raveen789
OP
9
Thanks for the help guys, its just the formula booklet does not indicate the 1/k ln|seckx| like it does for other formulas :/
Original post by raveen789
Thanks for the help guys, its just the formula booklet does not indicate the 1/k ln|seckx| like it does for other formulas :/


Yeah it seems fairly hit and miss. Just remember you must as it's 'the reverse chain rule'.

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