HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

[ayecee]

Re: MX2 Integration Marathon

 

[chris_power_96]

Re: MX2 Integration Marathon

View attachment 28661

For i,
-times both sides by cosec^2x
-use the substitution u=cotx
-Use partial fractions

for ii, use standard integral arctanx to get pi/4

 

[ayecee]

Re: MX2 Integration Marathon

For i,
-times both sides by cosec^2x
-use the substitution u=cotx
-Use partial fractions

for ii, use standard integral arctanx to get pi/4

Both correct, can you get (2)ii? the answer is 0.845.

 

[ayecee]

Re: MX2 Integration Marathon

Here's my attempt if it helps or if you spot a mistake. I don't have a worked solution, sorry

 

[Makematics]

Here's my attempt if it helps or if you spot a mistake. I don't have a worked solution, sorry
View attachment 28662

Im probs missing something, but to me it looks like you said sin (pi/2 -theta)=-cos(theta), which is incorrect

 

[Makematics]

For i,
-times both sides by cosec^2x
-use the substitution u=cotx
-Use partial fractions

for ii, use standard integral arctanx to get pi/4

Or you could multiply top and bottom by sec^2 theta and use the sub u=tanx which is simpler
for most people

 

[ayecee]

Re: MX2 Integration Marathon

Im probs missing something, but to me it looks like you said sin (pi/2 -theta)=-cos(theta), which is incorrect

Yes, you are correct, lol that mistake is embarrassing. Nevertheless, I made the mistake twice, from sin to cos, and from cos to sin, and in effect it made no difference once I reached the end, I still had pi/4[pi/4-pi/4].

Or you could multiply top and bottom by sec^2 theta and use the sub u=tanx which is simpler
for most people

That's what I did

 

[chris_power_96]

Re: MX2 Integration Marathon

Yes, you are correct, lol that mistake is embarrassing. Nevertheless, I made the mistake twice, from sin to cos, and from cos to sin, and in effect it made no difference once I reached the end, I still had pi/4[pi/4-pi/4].


That's what I did

When your integrating you cant use that rule twice, as it will always lead you to having a positive integral on both sides, meaning youre just solving 0=0, so i think somewhere you made a mistake in which it shouldn't go to 2I, but rather go to 0 on the lhs

 

[ayecee]

Re: MX2 Integration Marathon

When your integrating you cant use that rule twice, as it will always lead you to having a positive integral on both sides, meaning youre just solving 0=0, so i think somewhere you made a mistake in which it shouldn't go to 2I, but rather go to 0 on the lhs

Oh, I didn't realise that, although now I think about it it makes perfect sense; and indeed you are correct, I get an I on both sides... Looks like it must involve another method...

 

[Makematics]

Oh, I didn't realise that, although now I think about it it makes perfect sense; and indeed you are correct, I get an I on both sides... Looks like it must involve another method...

I tried investigating an IBP but it didnt work

 

[ayecee]

Re: MX2 Integration Marathon

I tried investigating an IBP but it didnt work

:/ I found a vey similar question in an old Coroneus textbook I have. The integral was the same, except the limits were from 0 to pi, where my method would work. The question I posted however has a previous part, so I guess it must be used somewhere in the solution. The one in the textbook also ha a previous part, asking you to prove the (a-x) relationship.
So maybe were missing something, or maybe it uses a simar rule...

 

[seanieg89]

Re: MX2 Integration Marathon

What is the source of the question as posted here?

 

[seanieg89]

Re: MX2 Integration Marathon

Actually dw, was about to suggest the possibility of a typo...but IBP does work.

 

[ayecee]

Re: MX2 Integration Marathon

What is the source of the question as posted here?

I got the question from some old handwritten questions my tutor gave me, and by old i mean 20 years old. I suspect he got the question from that old text book, but modified the question slightly.

Actually dw, was about to suggest the possibility of a typo...but IBP does work.

Righto, I'll have to give that a go then thanks

 

[Makematics]

Actually dw, was about to suggest the possibility of a typo...but IBP does work.

Ah ok, i probably used the wrong u and v in my IBP

 

[ayecee]

Re: MX2 Integration Marathon

Ah ok, i probably used the wrong u and v in my IBP

:/ I've tried usind u as x, sinx/(1+cosx^2), and as xsinx; each with no success.

 

[seanieg89]

Re: MX2 Integration Marathon

Upon actually doing the IBP working, the cancellation I expected did not occur... sorry about that. It may still be possible to evaluate this integral, but it is not obvious to me whether or not this can be done. As the integral to pi certainly can be done, I think this might be a typo / mistake of your tutor (how likely this is of course depends on how reputable your tutor is).

In any case, I wouldn't spend more time on it without seeing the question somewhere else or getting confirmation that it can be done by someone whose exact answer matches up with wolframalpha's numerical one.

 

[Sy123]

Re: MX2 Integration Marathon

 

[asianese]

Re: MX2 Integration Marathon

 

[Sy123]

Re: MX2 Integration Marathon

Yep

(but a=0 is also a solution )