Author Topic: Interesting questions (spesh) (Read 9282 times) Tweet Share

Martoman · « **on:** September 27, 2010, 08:44:44 pm »

So I did a post in methods, so I thought might as well in spesh. Anything that catches my eye I'll post here.

Exam 2 (only because this can't be done on a calc

) This is very difficult. A Kilbaha question for sure on 2011.
$\int \frac{1}{x} \sqrt{\frac{1-ln(x)}{1+ln(x)}}dx$ potter solved first

$\int \frac{x^2}{\sqrt{1-x^2}}dx$ an inspired substitution is needed for those who aren't doing unimaths (ie most speshies

) potter

$\int \frac{\sqrt{x+4}}{x}dx$ darksky

$\int \frac{1}{e^{3x}+4}dx$ chavi after direction :smitten:

$\int \frac{ln(cot(x))}{esin(2x)}dx$

$\int \frac{sin(x)cos(x)}{sin^{4}(x)+cos^4(x)}dx$

$\int \frac{sec(x)cos(2x)}{sin(x)+sec(x)}dx$

A nod to TSSM 09 -> $\int \frac{1+ lnx}{xlnx(1-lnx)}dx$

An extrememly MEAN possiblity for Exams this year

Find $\frac{d}{dx}(xln(x))$

Using these results find $\int \frac{xe^x}{\sqrt{1+e^x}}dx$

Potter · « **Reply #1 on:** September 27, 2010, 08:56:02 pm »

Sin-1(ln(x)) + [-(ln(x))^2 + 1]^1/2

Not sure if that's right..

wait.. Mucked up the signs.

Martoman · « **Reply #2 on:** September 27, 2010, 09:00:38 pm »

Almost, you forgot a root over the the 1-(ln(x))^2

Potter · « **Reply #3 on:** September 27, 2010, 09:02:05 pm »

Ohhh ooops. That's what you get for skipping steps. lol

Martoman · « **Reply #4 on:** September 27, 2010, 09:04:31 pm »

mmm I enjoyed this problem.

Potter · « **Reply #5 on:** September 27, 2010, 09:10:34 pm »

Yeah.. It's a nice little problem. I think the most difficult part is deciding what to do after the substitution. Where did you find it?

Martoman · « **Reply #6 on:** September 27, 2010, 09:13:42 pm »

This was stewart calculus I believe

Potter · « **Reply #7 on:** September 27, 2010, 09:22:53 pm »

Now.. This one I'm not sure about at all, but what I did seems logical.
1/2Sin-1(x) - [x(1-x^2)^1/2]/2 + C

I substituted in sinxcosx. Should've been 2sinxcosx

Martoman · « **Reply #8 on:** September 27, 2010, 09:30:24 pm »

veryyyyyy close i think you did something wrong with the /4 on the right term

brightsky · « **Reply #9 on:** September 27, 2010, 09:40:49 pm »

Substitute $u = \log_e(x)$ , $du = \frac{1}{x} dx$

$\implies \int \sqrt{\frac{1 - u}{1 + u}} du$

$= \int \frac{1 - u}{\sqrt{1 - u^2}} du$

$= \int \left(\frac{1}{\sqrt{1 - u^2}} - \frac{u}{\sqrt{1 - u^2}} \right) du$

Integrate term by term:

$= \arcsin(u) - \int \frac{u}{\sqrt{1 - u^2}} du$

Let $v = 1 - u^2$ , $dv = 2u du$

$\implies \arcsin(u) -\frac{1}{2} \int \frac{1}{\sqrt{v}} dv$

$= \arcsin(u) +\frac{2\sqrt{v}}{2}$

$= \arcsin(u) + \sqrt{v}$

Substitute $u = \log_e(x)$ and $v = 1 - u^2 = 1 - \log_e^2(x)$

Then $\implies \arcsin(\log_e(x)) + \sqrt{1 - \log_e^2(x)} + C$

Martoman · « **Reply #10 on:** September 27, 2010, 09:42:10 pm »

didn't have to ruin it for everyone but yes

brightsky · « **Reply #11 on:** September 27, 2010, 09:58:17 pm »

Oohh..ok SPOILER WARNING (dunno how to change colour of latex)...:p
Martoman can you check?

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Let $x = \sin(u)$ , $dx = \cos(u) du$

Then $\implies \int \frac{\sin^2(u)\cos(u)}{\sqrt{1 - \sin^2(u)}} du$

$= \int \frac{\sin^2(u) \cos(u)}{\sqrt{\cos^2(u)}} du$

$= \int \sin^2(u) du$

$= \int \left(\frac{e^{iu} - e^{-iu}}{2i}\right)^2 du$

$= \int \frac{e^{2iu} + e^{-2iu} -2}{4i^2} du$

$= \int \left(-\frac{e^{2iu} + e^{-2iu}}{4} + \frac{1}{2}\right) du$

$= \int \left(\frac{1}{2}\cos(2u) + \frac{1}{2}\right) du$

Integrate term by term:

$-\frac{1}{4} \sin(2u) + \frac{1}{2} u + C$

Substitute $u = \arcsin(x)$

$-\frac{1}{4} \sin(2\arcsin(x)) + \frac{1}{2} \arcsin(x) + C$

Martoman · « **Reply #12 on:** September 27, 2010, 10:01:52 pm »

yep thats right.... was eulers necessary?

brightsky · « **Reply #13 on:** September 27, 2010, 10:10:06 pm »

Substitute $u = \sqrt{x + 4}$ , $du = \frac{1}{2u} dx$

$\implies \int \frac{2u^2}{u^2 - 4} du$

$= 2 \int \frac{u^2}{u^2 - 4} du$

For the integrand $\frac{u^2}{u^2 - 4}$ :

$= 1 + \frac{4}{(u-2)(u+2)}$

$= 1 + \frac{1}{u-2} - \frac{1}{u+2}$

So $2 \int \left(1 + \frac{1}{u-2} - \frac{1}{u+2}\right) du$

Integrate term by term:

$2 (u + \log_e|u-2| - \log_e|u + 2|)$

Substitute $u = \sqrt{x+4}$

$2 (\sqrt{x + 4} + \log_e|\sqrt{x+4} - 2| - \log_e|\sqrt{x+4} + 2| + C$

EDIT: Woops. Cheers Martoman.

brightsky · « **Reply #14 on:** September 27, 2010, 10:10:39 pm »

Quote from: Martoman on September 27, 2010, 10:01:52 pm

yep thats right.... was eulers necessary?

No, but I suck at trig rules. :p Euler takes care of everything.

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