The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

C3 integration

Hi,

Please could someone check my working out? It isn't giving me the correct answer.

Integrate:∫sinxcos4x\displaystyle \int \frac{sinx}{cos^4x}

So this is what I did:

u=cosx→dudx=−sinx→dx=du−sinx u=cosx \rightarrow \frac{du}{dx}=-sinx \rightarrow dx=\frac{du}{-sinx}


∫sinxcos4xdu−sinx→−∫1cos4xdu→−∫u−4\displaystyle \int \frac{sinx}{cos^4x} \frac{du}{-sinx} \rightarrow -\displaystyle \int \frac{1}{cos^4x} du \rightarrow -\displaystyle \int u^{-4}

(u−33+c)→13cos3x+c (\frac {u^{-3}}{3} + c) \rightarrow \frac{1}{3cos^3x} + c

Would someone be able point out where I went wrong?

Thanks
Marc.
I actually can't spot the mistake xD what's the right answer>? :/
Reply 2
Avatar for Munrot07
Munrot07
18
Original post by marcsaccount
Hi,

Please could someone check my working out? It isn't giving me the correct answer.

Integrate:∫sinxcos4x\displaystyle \int \frac{sinx}{cos^4x}

So this is what I did:

u=cosx→dudx=−sinx→dx=du−sinx u=cosx \rightarrow \frac{du}{dx}=-sinx \rightarrow dx=\frac{du}{-sinx}


∫sinxcos4xdu−sinx→−∫1cos4xdu→−∫u−4\displaystyle \int \frac{sinx}{cos^4x} \frac{du}{-sinx} \rightarrow -\displaystyle \int \frac{1}{cos^4x} du \rightarrow -\displaystyle \int u^{-4}

(u−33+c)→13cos3x+c (\frac {u^{-3}}{3} + c) \rightarrow \frac{1}{3cos^3x} + c

Would someone be able point out where I went wrong?

Thanks
Marc.


Thats correct :smile:
Reply 3
Avatar for marcsaccount
marcsaccount
OP
5
thanks for confirming it for me guys.

I've left home now so haven't got access to what the book has for an answer, but there are a lot of errors in the book!

Thanks again.
Marc
Reply 4
Avatar for TenOfThem
TenOfThem
18
Original post by marcsaccount
Hi,

Please could someone check my working out? It isn't giving me the correct answer.

Integrate:∫sinxcos4x\displaystyle \int \frac{sinx}{cos^4x}

So this is what I did:

u=cosx→dudx=−sinx→dx=du−sinx u=cosx \rightarrow \frac{du}{dx}=-sinx \rightarrow dx=\frac{du}{-sinx}


∫sinxcos4xdu−sinx→−∫1cos4xdu→−∫u−4\displaystyle \int \frac{sinx}{cos^4x} \frac{du}{-sinx} \rightarrow -\displaystyle \int \frac{1}{cos^4x} du \rightarrow -\displaystyle \int u^{-4}

(u−33+c)→13cos3x+c (\frac {u^{-3}}{3} + c) \rightarrow \frac{1}{3cos^3x} + c

Would someone be able point out where I went wrong?

Thanks
Marc.



Did the book give the answer as 13sec⁡3x\frac{1}{3}\sec ^3x
Reply 5
Avatar for marcsaccount
marcsaccount
OP
5
Original post by TenOfThem
Did the book give the answer as 13sec⁡3x\frac{1}{3}\sec ^3x



The answer in the book is 18cos3x+c \frac{1}{8cos^3x} + c but there are many errors in it.
Original post by marcsaccount

Would someone be able point out where I went wrong?

Thanks
Marc.

Not sure why you'd do that.
That's the same as ∫tanx×sec3xdx\displaystyle \int tanx \times sec^3x dx which is a trivial integral.
Original post by marcsaccount
The answer in the book is 18cos3x+c \frac{1}{8cos^3x} + c but there are many errors in it.

Not correct. Don't think it is anyway.
(edited 10 years ago)
Reply 7
Avatar for davros
davros
16
Original post by keromedic
Not sure why you'd do that.



It's inspection or substitution - compare with the derivative of cosnxcos^n x
Original post by davros
It's inspection or substitution - compare with the derivative of cosnxcos^n x

I think I'm missing the point you're making.
Unparseable latex formula:

\dfrac{d}{dx}cos^nx= -nsinxcos^{n-1}x{?}=\dfrac{-nsin2x \timtes cos^{n-2}}{2}{?}

(edited 10 years ago)
Reply 9
Avatar for TenOfThem
TenOfThem
18
Original post by keromedic
I think I'm missing the point you're making.
Unparseable latex formula:

\dfrac{d}{dx}= -nsinxcos^{n-1}x{?}=\dfrac{-nsin2x \timtes cos^{n-2}}{2}{?}



I think you are

ddxcos⁡nx=−nsin⁡xcos⁡n−1x\dfrac{d}{dx} \cos ^nx = -n\sin x\cos ^{n-1}x

Since the integrand is sin⁡xcos⁡−4x\sin x\cos ^{-4}x

So n = -3

And there needs to be a division by 3
(edited 10 years ago)
Original post by TenOfThem
I think you are

ddxcos⁡nx=−nsin⁡xcos⁡n−1x\dfrac{d}{dx} \cos ^nx = -n\sin x\cos ^{n-1}x

Since the integrand is sin⁡xcos⁡−4x\sin x\cos ^{-4}x

So n = -3

And there needs to be a division by 3

Oh ofcourse :facepalm:

Thanks :biggrin: :h:.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you