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Integrate 1 + sinx/cos x

I'm quite stuck on this question: Integrate 1 + sinx/cosx by substitution.
The book also says let u=sin x

Now I was thinking of changing it into sec x + tan x but that didn't get me far

Should I bring cosx to the top?

Sorry btw, I can't use latex so please bear with me!

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Reply 1
If you substitute u=sinxu=\sin x then you get dx=ducosxdx = \dfrac{du}{\cos x}, so you get cos2x\cos^2 x on the denominator. How might you write cos2x\cos^2 x in terms of uu?
(edited 11 years ago)
Reply 2
Original post by nuodai
If you substitute u=cosxu=\cos x then you get dx=ducosxdx = \dfrac{du}{\cos x}, so you get cos2x\cos^2 x on the denominator. How might you write cos2x\cos^2 x in terms of uu?


U^2

e: i got negged coz i got it wrong :facepalm2:
(edited 11 years ago)
Reply 3
So I got 1 + sinx du/cos^2x?
Reply 4
Original post by This Honest
U^2


No. u=sinxu=\sin x so u2=sin2xcos2xu^2 = \sin^2 x \ne \cos^2 x - try again!

Original post by This Honest
So I got 1 + sinx du/cos^2x?


Not quite, you should have 1+sinxcos2xdu\dfrac{1 + \sin x}{\cos^2 x} du. Now put this in terms of uu.
Reply 5
Original post by This Honest
So I got 1 + sinx du/cos^2x?


Then use the fact that u= sinx

and cos^2x = 1-sin^2x.
Reply 6
Original post by nuodai
No. u=sinxu=\sin x so u2=sin2xcos2xu^2 = \sin^2 x \ne \cos^2 x - try again!



Not quite, you should have 1+sinxcos2xdu\dfrac{1 + \sin x}{\cos^2 x} du. Now put this in terms of uu.



Original post by f1mad
Then use the fact that u= sinx

and cos^2x = 1-sin^2x.


Alright guys,

I've now got: 1 + u/1-sin^x
(edited 11 years ago)
Reply 7
Original post by This Honest
Alright guys,

I've now got: 1 + u/1-sin^x


Write it in terms of uu; that still has an xx in it. You'll notice that you can factorize something that simplifies the expression into something you can integrate.
Reply 8
Original post by This Honest
I'm quite stuck on this question: Integrate 1 + sinx/cosx by substitution.
The book also says let u=sin x

Now I was thinking of changing it into sec x + tan x but that didn't get me far

Should I bring cosx to the top?

Sorry btw, I can't use latex so please bear with me!


This question can also be easily done without substitution,

(1+sinxcosx)dx \displaystyle \int \left(1+\frac{sinx}{cosx}\right) dx

Now try to differentiate, u=ln(cosx) \displaystyle u=ln(cosx)
Reply 9
integrate: 1 +u/1-u^2 ??
Original post by This Honest
I'm quite stuck on this question: Integrate 1 + sinx/cosx by substitution.
The book also says let u=sin x

Now I was thinking of changing it into sec x + tan x but that didn't get me far

Should I bring cosx to the top?

Sorry btw, I can't use latex so please bear with me!


I'm sorry but i'm too busy to help you
Original post by This Honest
Alright guys,

I've now got: 1 + u/1-sin^x


1+u/1-u^2

remember that u = sinx and that cos^2(x) = 1 - sin^2(x)
Reply 12
Original post by nuodai
Write it in terms of uu; that still has an xx in it. You'll notice that you can factorize something that simplifies the expression into something you can integrate.


:woo: 1/1-u
Reply 13
Original post by This Honest
integrate: 1 +u/1-u^2 ??


Yes, remember you will have to use partial fraction.

1+u1u2=1+u(1u)(1+u) \displaystyle 1 + \frac{u}{1-u^2} = 1 + \frac{u}{(1-u)(1+u)}
Reply 14
I GOT IT!

-ln!1-sinx! + c

can't find modular signs on keyboard :colondollar:$

Thanks everyone for the help
Original post by This Honest
integrate: 1 +u/1-u^2 ??


have you learnt the integration rule, where if you differentiate the denominator, you get the value that appears in the numerator. hence ln(your original function)

have you come across this before?

this could prove useful here.

someone else correct me if im wrong

WHOOPS PARTIAL FRACTIONS ARE THE WAY FORWARD :woo:
Reply 16
Original post by raheem94
Yes, remember you will have to use partial fraction.

1+u1u2=1+u(1u)(1+u) \displaystyle 1 + \frac{u}{1-u^2} = 1 + \frac{u}{(1-u)(1+u)}


I didn't quite understand your method but thanks anyways :smile:
Reply 17
Original post by blacklistmember
I'm sorry but i'm too busy to help you


Why post then?
Reply 18
Original post by James A
have you learnt the integration rule, where if you differentiate the denominator, you get the value that appears in the numerator. hence ln(your original function)

have you come across this before?

this could prove useful here.

someone else correct me if im wrong

WHOOPS PARTIAL FRACTIONS ARE THE WAY FORWARD :woo:


OHHHHHHH YEAHHHHHHH!

The book said "by sub" so my mind was only on sub and nothing else :colonhash:
Reply 19
Original post by raheem94
Yes, remember you will have to use partial fraction.

1+u1u2=1+u(1u)(1+u) \displaystyle 1 + \frac{u}{1-u^2} = 1 + \frac{u}{(1-u)(1+u)}


I don't think so.

If it's (1+u)/ (1-u^2) = (1+u)/(1-u)(1+u) = 1/(1-u) = - I of -1/(1-u) then it's a simple f'(u)/f(u) integral.

I think you mis-interpreted :tongue:.

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