The Student Room Group

Integrate 1 + sinx/cos x

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Reply 20
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:


Look at post no. 9, i have hinted towards this method.
Original post by blacklistmember
I'm sorry but i'm too busy to help you


you wally.
Original post by f1mad
Why post then?


no idea
Reply 23
Original post by James A

WHOOPS PARTIAL FRACTIONS ARE THE WAY FORWARD :woo:


It's not.
Original post by raheem94
Look at post no. 9, i have hinted towards this method.


yeah i know, but i forgot to include your post in my response. Sorry.
Original post by James A
you wally.


i had to look that up :P
Original post by f1mad
It's not.


:colone:
Reply 27
Original post by f1mad
It's not.


Both ways, can be used. Your method is faster, but doing partial fractions also takes moments.
Reply 28
Original post by This Honest
integrate: 1 +u/1-u^2 ??

Please put in brackets with things like this, i.e. (1+u)/(1-u²). Not doing so has led to some misleading advice because people think you mean 1+u1u21+\dfrac{u}{1-u^2} and 1+sinxcosx1 + \dfrac{\sin x}{\cos x} instead of 1+u1u2\dfrac{1+u}{1-u^2} and 1+sinxcosx\dfrac{1+\sin x}{\cos x}.

Anyway, you got the right answer ln1sinx+C-\ln \left| 1 - \sin x \right| + C.

(In fact you can remove the mod signs since 1sinx01-\sin x \ge 0 for all xx.)
Reply 29
Original post by raheem94
Both ways, can be used. Your method is faster, but doing partial fractions also takes moments.


I beg to differ: your partial fraction isn't the same as the function of the substitution (see Nuodai's post).
Reply 30
Original post by nuodai
Please put in brackets with things like this, i.e. (1+u)/(1-u²). Not doing so has led to some misleading advice because people think you mean 1+u1u21+\dfrac{u}{1-u^2} and 1+sinxcosx1 + \dfrac{\sin x}{\cos x} instead of 1+u1u2\dfrac{1+u}{1-u^2} and 1+sinxcosx\dfrac{1+\sin x}{\cos x}.

Anyway, you got the right answer ln1sinx+C-\ln \left| 1 - \sin x \right| + C.

(In fact you can remove the mod signs since 1sinx01-\sin x \ge 0 for all xx.)


Sound like my maths teacher now :tongue:
Thanks, I will next time :yep:
Reply 31
Original post by f1mad
I beg to differ: your partial fraction isn't the same as the function of the substitution (see Nuodai's post).


So we were solving the wrong question. The question was different, right?
Reply 32
Original post by This Honest
Sound like my maths teacher now :tongue:
Thanks, I will next time :yep:


I've been made grumpy with age :p:
Reply 33
Original post by raheem94
So we were solving the wrong question. The question was different, right?


Yeah..sorry that was my fault. As noudai said...I caused confusion in the way I wrote it out :colondollar:
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?


Why have you made u equal to cos x here?
Reply 35
Original post by This Honest
Yeah..sorry that was my fault. As noudai said...I caused confusion in the way I wrote it out :colondollar:


No problem, but please learn LaTeX, it only takes a few minutes to learn it.
Reply 36
Original post by raheem94
No problem, but please learn LaTeX, it only takes a few minutes to learn it.


I'll try and learn it when I'm having difficulty with C4 again before I post the problem :smile:
Reply 37
Original post by TheGrinningSkull
Why have you made u equal to cos x here?


Typo; I'd set u=sinxu=\sin x. If I'd actually set u=cosxu=\cos x then I wouldn't have got dx=ducosxdx = \dfrac{du}{\cos x}, after all.
An alternative approach would be to notice that 1+sinxcosx1+cos(π2x)sin(π2x)2cos2(π4x2)2sin(π4x2)cos(π4x2)\dfrac{1+\sin x}{\cos x} \equiv \dfrac{1+\cos (\frac{\pi}{2} -x)}{\sin (\frac{\pi}{2} - x)} \equiv \dfrac{2\cos^2 (\frac{\pi}{4} - \frac{x}{2})}{2\sin (\frac{\pi}{4} - \frac{x}{2})\cos (\frac{\pi}{4} - \frac{x}{2})}, which after a bit of cancelling leads to logs directly.

EDIT: just realised the OP was asked for a specific sub, so this can be ignored.
(edited 11 years ago)
Reply 39
Original post by Farhan.Hanif93
An alternative approach would be to notice that 1+sinxcosx1+cos(π2x)sin(π2x)2cos2(π4x2)2sin(π4x2)cos(π4x2)\dfrac{1+\sin x}{\cos x} \equiv \dfrac{1+\cos (\frac{\pi}{2} -x)}{\sin (\frac{\pi}{2} - x)} \equiv \dfrac{2\cos^2 (\frac{\pi}{4} - \frac{x}{2})}{2\sin (\frac{\pi}{4} - \frac{x}{2})\cos (\frac{\pi}{4} - \frac{x}{2})}, which after a bit of cancelling leads to logs directly.


:cry2: thanks for that.
I'd stick with my method though :tongue:

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