How Good Am I at Integrals?

[Gib Z]

Ok well that gets me to

$$\frac{\pi}{4} \log_e \sqrt{2} - \frac{\sqrt{2}}{2} - \int^{\pi/4}_0 \log_e \cos (\pi/4 - t) dt$$

Wait no it doesn't..i give up

 

[morphism]

$$\int_0^{\pi/4} \ln \left( \cos \left( t - \frac{\pi}{4} \right) \right) \; dt - \int_0^{\pi/4} \ln(\cos t) \; dt = 0$$

 

[Gib Z]

>.<" Sorry I must be really annoying and annoyed today, I got the first bit down to be the same as the second integral, but for some reason, ignoring that I used the same bounds property on the 2nd integral one as well. ~sigh~

 

[morphism]

Made any progress with the other one? It's really slick. Possibly the most amusing of all the ones I posted.

 

[Gib Z]

Amusing? I swear you work for the CIA, testing your torture methods on us!

 

[Gib Z]

Ok I did the cheapskates desperate way out!

Since log (1+x) has a nice Taylor series, I expanded the Taylor series and integrated term by term. Once I did this I noticed the antiderivatve as the Polylogarithm function.

So I get that the integral must evaluate to $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2}$$. That looks like an alternating version of the sum of the reciprols of the squares, which i know is $$\pi^2/6$$. Thats my best.

 

[morphism]

Don't give up yet! That's the way I did it as well. Try to find a closed form for that sum using $\sum 1/n^2 = \pi^2/6$. (Hint: What happens when you subtract one from the other?)

 

[Gib Z]

$$\frac{\pi^2}{12}$$. Edit: Sorry I took so long to reply on this one, was a bit busy. I can't believe I actually got that far, and then missed out on how to finish it. I loved ur ones morphism, keep em coming.

 

[TripleS]

dx/(4+x^2)

(sin^-1 x + cos^-1 x) dx

 

[Gib Z]

$$\int \frac{1}{x^2+2^2} dx$$

Simple, let x=2 tan theta. Then dx=2 sec^2 theta
$$\int \frac{2\sec^2 \theta}{4\tan^2 \theta + 4} d\theta = \int \frac{2\sec^2 \theta}{4(\tan^2 \theta + 1)} d\theta = \int \frac{2\sec^2 \theta}{4\sec^2 \theta} d\theta = \int \frac{1}{2}d\theta = \frac{\theta}{2} +C$$.

Since tan theta = x/2, theta = arctan (x/2)

$$\int \frac{1}{x^2+4}dx = \frac{1}{2}\arctan(x/2) +C$$.

$$\int \arcsin x + \arccos x dx = \frac{\pi x}{2}$$

I did that in my head because I know the integrand is a constant pi/2, but substitution u = arcsin x + arccos x works as well.

 

[VietDao29]

And, how about these:
Some simple little nice ones. :)
I like the second best. :!)

1. $$\int \frac{dx}{\cos x (\sin x + \cos x)}$$

2. $$\int \frac{(x + 7) ^ 5}{(x + 2) ^ 7} dx$$

3. $$\frac{5}{x (x ^ {2000} + 1)} dx$$

 

[Gib Z]

:'( I can't seem to get the last 2...

1.$$\int \frac{dx}{\cos x (\sin x + \cos x)} = \int \frac{dx}{\cos^2 x (\tan x + 1)} = \int \frac{\sec^2 x}{\tan x + 1} dx = \log_e |\tan x +1|$$

 

[TripleS]

fix the coding
[ /tex]

 

[Gib Z]

Yea I Did already lol

EDIT: O btw You aren't on msn, so if your here, we going to play basketball at strathfield park by like 1:30, come.

 

[TripleS]

yeps

how do u work out the primitive of cosec^2 x?

and one for you dx/ (sqrt (x-x^2))

 

[Gib Z]

1. The cosec^2 x one.

$\sin^2 x + \cos^2 x =1$
$1+\cot^2 x = \csc^2 x$

Therefore
$$\int \csc^2 x dx = \int ( \cot^2 x+ 1) dx = \int \frac{\cos x}{\sin x} dx + \int 1 dx$$

1st Integral, u = sin x, then its easy.
2nd Integral is trivial.

$$\int \csc^2 x = \log_e |\sin x| + x$$
For the 2nd one, u = sqrt ( x- x^2), I end up getting

$$\frac{ 2\sqrt{x-x^2} \sqrt{x} \log_e (\sqrt{x-1} + \sqrt{x})}{\sqrt{x-x^2}}$$.

I Differentiated it to check, I am right :) Why don't you differentiate it for me again?

BTW Hams says : you suck, stop wasting your life on maths like ragib =P

DONT LISTEN TO HIM!

 

[mathPimpDaddy]

is # 2 like extremely long? Very clever Gib z

 

[TripleS]

1. The cosec^2 x one.

$\sin^2 x + \cos^2 x =1$
$1+\cot^2 x = \csc^2 x$

Therefore
$$\int \csc^2 x dx = \int ( \cot^2 x+ 1) dx = \int \frac{\cos x}{\sin x} dx + \int 1 dx$$

1st Integral, u = sin x, then its easy.
2nd Integral is trivial.

$$\int \csc^2 x = \log_e |\sin x| + x$$



For the 2nd one, u = sqrt ( x- x^2), I end up getting

$$\frac{ 2\sqrt{x-x^2} \sqrt{x} \log_e (\sqrt{x-1} + \sqrt{x})}{\sqrt{x-x^2}}$$.

I Differentiated it to check, I am right :) Why don't you differentiate it for me again?

BTW Hams says : you suck, stop wasting your life on maths like ragib =P

DONT LISTEN TO HIM!

ham is cool

i never thought the second one was that confusing
can u do it with working ..please

 

[VietDao29]

...
and one for you dx/ (sqrt (x-x^2))

Err... you seemed to have messed up the second one a bit.

$$\int \frac{dx}{\sqrt{x - x ^ 2}} = \int \frac{dx}{\sqrt{\frac{1}{4} - \left( x - \frac{1}{2} \right) ^ 2}} = \arcsin \left[\frac{\left( x - \frac{1}{2} \right)}{\frac{1}{2}} \right] + C = \arcsin (2x - 1) + C$$

$$\int \csc^2 x = \log_e |\sin x| + x$$

Uhm... I think it should have read:
$$\int \csc ^ 2 x dx = - \cot x + C$$

-----------------------

IMHO, the second one is a nice one. It goes like this:
$$\int \frac{(x + 7) ^ 5}{(x + 2) ^ 7} dx = \int \frac{(x + 7) ^ 5}{(x + 2) ^ 5 (x + 2) ^ 2} dx = \int \left( \frac{x + 7}{x + 2} \right) ^ 5 \frac{1}{(x + 2) ^ 2} dx = ...$$
From here, a proper u-substitution should work.

 

[Gib Z]

Please Shoot me..eek Your right of course..I stuffed up both..and what's worse is that I Knew what I was going to do, since TripleS wanted to know how to get the integral, rather than Just know it. However this is just one of the basic ones you should just be able to do straight, like the primitive of x^2. I will post you some work, but you will see it is very circular.

As You can see, I forgot a square, and I continued as if there was no square.
Heres a correction:
$$\int \csc^2 x dx = \int ( \cot^2 x+ 1) dx = \int (\frac{\cos x}{\sin x})^2 dx + \int 1 dx = \int \frac{\cos x\sqrt{1-\sin^2 x}}{\sin^2 x} dx + x$$

For the remaining integral, u = sin x reduces it to
$$\int \frac{\sqrt{1-u^2}}{u^2} du$$

Now The problem with this is that you have to use a trig substitution that will get you back to the original problem. Trig substitution helps when you know how to evaluate the remaining trig integral but in this case it is the original integral. Dang.Ahh Thats A good one VietDao29, u = x+7 over x+2, du = -5 dx over (x+2)^2

So that makes it simple, I get $$\frac{(\frac{x+7}{x+2})^6}{-30}$$.
For yours 3rd one I tried partial fractions which don't work..and Integration by parts..nope..and some Algebraic Manipulations and none of those worked...

 

[mathPimpDaddy]

#3 partial fractions is:

5/x - 5x^1999/(x^2000 + 1) right? Then the integral should be easy.

 

[Gib Z]

How did you do the partial fractions ?

When doing partial fractions don't you have to have to have all the terms of the polynomial. Eg For 1/x(x^5+4) shouldn't it be A/ x + (Bx^4 + cx^3 + dx^2...)/x^5 +4?

 

[Gib Z]

Well anyway Yup Now I got the integral: $5\ln |x| - \frac{\ln (x^{2000}+1)}{400}$

 

[mathPimpDaddy]

And, how about these:
Some simple little nice ones. :)
I like the second best. :!)

1. $$\int \frac{dx}{\cos x (\sin x + \cos x)}$$

2. $$\int \frac{(x + 7) ^ 5}{(x + 2) ^ 7} dx$$

3. $$\frac{5}{x (x ^ {2000} + 1)} dx$$

I very much liked the 2nd one Viet! Very good problems though, you learn a lot from them.

 

[VietDao29]

#3 partial fractions is:

5/x - 5x^1999/(x^2000 + 1) right? Then the integral should be easy.

Yes, the third problem is:
$$\int \frac{5}{x (x ^ {2000} + 1)} dx = 5 \int \frac{x ^ {2000} + 1 - x ^ {2000}}{x (x ^ {2000} + 1)} dx$$
$$= 5 \int \left( \frac{1}{x} - \frac{x ^ {1999}}{x ^ {2000} + 1} \right) dx = 5 \left( \ln|x| - \frac{1}{2000} \ln (x ^ {2000} + 1) \right) + C$$

Well anyway Yup Now I got the integral: $5\ln |x| - \frac{\ln (x^{2000}+1)}{400}$

This is correct.

...So that makes it simple, I get $$\frac{\left( \frac{x + 7}{x + 2} \right) ^ 6}{-30}$$

This is also correct.

 

[Gib Z]

Hey guys! Some people in the homework section wanted this thread revived, but with a new intent.

The original purpose was for people to post up integrals (usually indefinite) for me to solve. However the renewed purpose is for anyone to post up a particularly difficult problem for anyone to solve. The problems should be able to be worked out with no more than CalcII knowledge please. It does not necessarily have to be an integral, it could be an interesting piece of number theory or the computation of the value of a series, or even a particularly difficult exercise of algebraic manipulation!

It would be wonderful if you all would participate :)

It has concerned some that this thread will be exploited by some for their homework to be solved, and a way to avoid this would be extra work on part of the mentors. For this I propose that Only members with post counts of over 500, or members who post solutions to the previous problem, will be allowed to post new questions. Generally I have found people with higher post counts know quite a lot and are more accustomed to the forums rules, so will not exploit this. And granted members can solve the previous problem, they probably would not need help with their homework :)

I think I will go first :) (My Class is learning the compound interest formula, this is definitely not my homework):

Evaluate $$\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^m\left(m\,3^n+n\,3^m\right)}$$

This is quite hard.

 

[mathwonk]

how about problems for which one needs to set up the integral, like say, find the "volume" of the unit ball in 4 space?

(by the way this thread title reminds me of the old song "how dry i am")

 

[Parthalan]

Any excuse to use hyperspherical coordinates (I think) will do!

 

[Gib Z]

Hyperspherical co ordinates and 4 space doesn't sound like CalcII level to me :( I guess if you wanted you could post such questions, but I definitely would not be able to participate :P

 

[yip]

In response to GibZ's symmetrical question:
$$\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^ m\left(m\,3^n+n\,3^m\right)}=\frac{1}{2}[\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^ m\left(m\,3^n+n\,3^m\right)}+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{n^2\,m}{3^ n\left(n\,3^m+m\,3^n\right)}]$$
$$=\frac{1}{2}[\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^{2}n.3^{n}(n3^{m}+m3^{n})+n^{2}m.3^{m}(m3^{n}+n3^{m})}{3^{m+n}(m3^{n}+n3^{m})^{2}}]$$
$$=\frac{1}{2}[\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{mn}{3^{m+n}}]=\frac{1}{2}\sum_{m=1}^\infty\frac{m}{3^{m}}\sum_{n=1}^\infty\frac{n}{3^{n}}$$
$$\sum_{m=1}^\infty\frac{m^{2}}{3^{m}}=(\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+...)$$
$$=(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...)+(\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...)+(\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+...)+...$$
$$=\frac{\frac{1}{3}+\frac{1}{9}+...}{\frac{2}{3}}=\frac{3}{4}$$
$$\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^ m\left(m\,3^n+n\,3^m\right)}=\frac{1}{2}\frac{9}{16}=\frac{9}{32}$$

 

[Gib Z]

yip, you are one smart cookie :) That is exactly correct, which means you get to post the next question if you want :)

I didn't think anyone would solve it that fast >.<

 

[yip]

Here is a rather tricky integral:

Integrate

$$\int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}}$$

 

[Gib Z]

Mathematica gives the answer is terms of various non-elementary functions...But I've done it before and I can do it again, I'm sure its much simpler than the Integrator says it is.

 

[Gib Z]

$$\int{\frac{{(1+x^{2})}}{{(1-x^{2})\sqrt{1+x^{4}}}}dx = }\int{\frac{{x^{2}(\frac{1}{{x^{2}}}+1)}}{{x^{2}(\frac{1}{x}-x)\sqrt{\frac{1}{{x^{2}}}+x^{2}}}}dx}$$

Let $${(\frac{1}{x}-x)}=t$$, then $${(\frac{1}{{x^{2}}}+1)dx}=-dt$$.

$$\int{\frac{{(1+x^{2})}}{{(1-x^{2})\sqrt{1+x^{4}}}}dx = }-\int{\frac{{dt}}{{t\sqrt{t^{2}+2}}}} = -\log \left({\frac{t}{{\sqrt{t^{2}+2}+\sqrt 2 }}}\right) + C$$

Rewrite in terms of x and we are done :)

 

[yip]

That seems to be correct, though I used the substitution t=x/(1-x^2), which works out a bit nicer