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The hard integral thread.

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Reply 580
Original post by TeeEm
there is a standard way of doing

(Acosx +Bsinx)/(Ccosx +Dsinx)


I know you can use a substitution to directly solve the integral however there is another nice way of doing it.

I haven't checked if the substitution works with a and b coefficients however it defiantly works for

cosxsinx+cosxdx \displaystyle\int \dfrac{\cos{x}}{\sin{x} + \cos{x}} \mathrm{d} x
Reply 581
Original post by Louisb19
I know you can use a substitution to directly solve the integral however there is another nice way of doing it.

I haven't checked if the substitution works with a and b coefficients however it defiantly works for

cosxsinx+cosxdx \displaystyle\int \dfrac{\cos{x}}{\sin{x} + \cos{x}} \mathrm{d} x



I do not think you need substitution for any of these.
Reply 582
Original post by TeeEm
I do not think you need substitution for any of these.


That is what I said, you can use a substitution although it doesn't have the beauty of the method I had in mind.

Could you elaborate on your method for me?
(edited 8 years ago)
Reply 583
Original post by Louisb19
That is what I said, you can use a substitution although it doesn't have the beauty of the method I had in mind.

Could you elaborate on your method for me?


look at IYGB special paper O, Q18
Reply 584
For AEAers or STEPers only
(just written and it is to be included into one of my papers)
Original post by TeeEm
For AEAers or STEPers only
(just written and it is to be included into one of my papers)


I think x=9-2cosu shud work.


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Reply 586
Original post by physicsmaths
I think x=9-2cosu shud work.


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I know what works and as this will go to my IYGB Specials it requires no Further maths.
Maybe it works also
try it
Original post by TeeEm
I know what works and as this will go to my IYGB Specials it requires no Further maths.
Maybe it works also
try it


Oh so you didn't use that sub? If not I'll try it now.


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Reply 588
Original post by physicsmaths
Oh so you didn't use that sub? If not I'll try it now.


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the sub is "special reserve" knowledge.
the answer is π - 2
Original post by TeeEm
the sub is "special reserve" knowledge.
the answer is π - 2


ImageUploadedByStudent Room1450382699.415058.jpg
In the 2nd line I multiply top and bottom by (1-cosu)^1/2

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(edited 8 years ago)
Reply 590
Original post by TeeEm
the sub is "special reserve" knowledge.
the answer is π - 2


I got pi + 2 - 2root(2)

Probably a silly mistake.

EDIT: Woops no I got pi - 2
(edited 8 years ago)
Reply 591
Original post by physicsmaths
ImageUploadedByStudent Room1450382699.415058.jpg
In the 2nd line I multiply top and bottom by (1-cosu)^1/2

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I will have to accept, you are multiplying top and bottom by zero, but this is a job for the purists to discuss ...
Reply 592
Original post by TeeEm
the sub is "special reserve" knowledge.
the answer is π - 2


My method was this.

Substitution of u2=x7 u^2 = x - 7

By parts using s=u s = u and dtdu=u(4u2)12 \dfrac{\mathrm{d} t}{\mathrm{d} u} = \dfrac{u}{(4 - u^2)^{\frac{1}{2}}}

Substitution of u=2sinθ u = 2 \sin{\theta}

Answer π2 \pi - 2

Granted it's probably not the most elegant method.
(edited 8 years ago)
Reply 593
Original post by Louisb19
My method was this.

Substitution of u2=x7 u^2 = x - 7

By parts using s=u s = u and dtdu=u(4u2)12 \dfrac{\mathrm{d} t}{\mathrm{d} u} = \dfrac{u}{(4 - u^2)^{\frac{1}{2}}}

Substitution of u=2sinθ u = 2 \sin{\theta}


Granted its probably not the most elegant method.


if it works it is good
Reply 594
Original post by physicsmaths

In the 2nd line I multiply top and bottom by (1-cosu)^1/2

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Those integral signs lmao.

Nice method. How did you spot that sub?
Original post by Louisb19
Those integral signs lmao.

Nice method. How did you spot that sub?

Not sure lol, i knew it would be some trig one. And just thought of the one that would work nicely to cancel something out.


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Reply 596
Original post by Louisb19
Those integral signs lmao.

Nice method. How did you spot that sub?


Original post by physicsmaths
Not sure lol, i knew it would be some trig one. And just thought of the one that would work nicely to cancel something out.


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Read carefully the comment in my solution and copy it into your "arsenal"
test it with another very easy one
Reply 597
Original post by TeeEm
Read carefully the comment in my solution and copy it into your "arsenal"
test it with another very easy one


I guess mine was a some what similar idea, no where near as efficient. Nice solution Tee.

If you wouldn't mind could you give me some advice? Do you have any tips on how to deal with exam anxiety, I tend to perform a lot worse when I'm under pressure and I'm scared that I will tank STEP despite putting in a lot of work this year. I know I'm not Zacken or anything but I really enjoy Maths, I want to prove to myself that I am good enough to get a good grade. I really don't want a repeat of MAT.

Thanks.
(edited 8 years ago)
Original post by Louisb19

If you wouldn't mind could you give me some advice? Do you have any tips on how to deal with exam anxiety, I tend to perform a lot worse when I'm under pressure and I'm scared that I will tank STEP despite putting in a lot of work this year. I know I'm not Zacken or anything but I really enjoy Maths, I want to prove to myself that I am good enough to get a good grade. I really don't want a repeat of MAT.


Looool, you should be glad you're not me, I'm nowhere near as good as the likes of Krollo or 16characters... :tongue: :rofl:
Original post by TeeEm
Read carefully the comment in my solution and copy it into your "arsenal"
test it with another very easy one


1011dx18x77x2=1011dx4(x9)2\displaystyle \int_{10}^{11} \frac{\mathrm{d}x}{\sqrt{18x - 77 - x^2}} = \int_{10}^{11} \frac{\mathrm{d}x}{\sqrt{4 - (x-9)^2}}

Which evaluates quite cleanly to [arcsin(x92)]1011=arcsin1arcsin12=π3\displaystyle \left[\arcsin \left(\frac{x-9}{2}\right)\right]_{10}^{11} = \arcsin 1 - \arcsin \frac{1}{2} = \frac{\pi}{3}

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