Are You Ready to Challenge Your Integral Solving Skills?

[Hootenanny]

Not technically difficult, but a nice introduction to recurrence methods;

I = \int{e^x\cdot\sin(x)}\; dx

 

[morphism]

\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \; dx

and

\int_0^{\pi / 2} \frac{1}{1 + \tan^{\pi} x} \; dx

are interesting.

 

[Eighty]

morphism: Neat, integrals I could do but not Mathematica. :)

 

[Gib Z]

Hootenanny - I = e^x \sin x - \int e^x \cos x dx
\int e^x \cos x dx = e^x \cos x + I
I= \frac{e^x(\sin x - \cos x)}{2}

morphism - Using the standard result \int^a_0 f(x) dx = \int^a_0 f(a-x), your first integral is shown below:

I= \int^{\pi}_0 \frac{x\sin x}{1+\cos^2 x}dx = \int^{\pi}_0 \frac{\pi \sin x}{1+\cos^2 x}dx - I

2I = \pi \int^{\pi}_0 \frac{\sin x}{1+\cos^2 x} dx

Can't be bothered typing up all the latex, but basically u=cos x, and arctan form gets us :
I = \frac{\pi^2}{4}.

As for your second one, I find your torture methods interesting. We have a vacancy at Guantanamo Bay if you would like to apply.

 

[morphism]

Hehe

(Your solution for the first one is correct of course. Good job!)

 

[murshid_islam]

how do i solve morphism's second one?
\int_0^{\pi / 2} \frac{1}{1 + \tan^{\pi} x} \; dx

 

[morphism]

Use the "standard result" Gib posted (i.e. essentially the substitution u = pi/2 - x). Then mess around with things.

 

[Hootenanny]

Okay, one more trig one. I haven't looked through your previous post so I apologise if this has already been posted;

I = \int{\frac{dx}{4+\sin(x)}}

 

[Gib Z]

Nvm Much easier than I thought. I didn't even give it a try because I seemed to think the exponent of pi was a bad sign.
Let I be the original Integral.
I= \int^{\pi/2}_0 \frac{1}{1+\tan^{\pi} x} dx = \int^{\pi/2}_0 \frac{1}{1+\cot^{\pi} x} dx.

2I = \int^{\pi/2}_0 \frac{1}{1+\tan^{\pi} x} dx + \int^{\pi/2}_0 \frac{1}{1+\cot^{\pi} x} dx = \int^{\pi/2}_0 \frac{1}{1+\cot^{\pi} x} + \frac{1}{1+\tan^{\pi} x} dx. Simple cross multiplying and expansion gets us 2I=\int^{\pi/2}_0 1 dx = \pi/2

Therefore I = \frac{\pi}{4}.

Nice ones morphism, got any more?

 

[Gib Z]

Umm Hootenannys one took a while, I Basically did t=tan (x/2) and made it a rational function so on so forth.

I got \frac{2}{\sqrt{15}} \arctan \frac{4 \tan (x/2) +1}{\sqrt{15}} + C which seems like a hairy answer, I don't think I am right. I can't be stuffed to differentiate it to check lol.

 

[morphism]

Nice ones morphism, got any more?

Here you go:

\int_0^{\pi / 2} \tan^{-1} x + \cot^{-1} x \; dx

These are inverses not reciprocals. (N.B.: There exists a solution that is one line long.)

\int_0^1 \frac{\log(x + 1)}{x^2 + 1} \; dx

\int_0^1 \frac{\log(x + 1)}{x} \; dx

\int_0^\pi \log (\sin x) \; dx

 

[Gib Z]

Hehe the first one is easier than I thought. I first tried u = arctan x + arccot x, but when I tried to find du I realized the derivative was 0, so I finally noticed it was a constant function. I tried x=1, the constant function is equal to pi/2, so the solution is \frac{\pi^2}{4}!

I couldn't do the 2nd one :(

Third one I did u= log (2), du/dx=1/x which is in there so all good.
So I integrated it and ended up with \frac{\log^2 (x+1)}{2}.

Last one, I split it up into 2 integrals, from 0 to pi/2, and pi/2 to pi.
Using symmetry arguments and the such, I got that it 2I= \int^{\pi/2}_0 \log(\sin x) dx = \int^{\pi/2}_0 \log(\cos x) dx but I am stuck from there.

 

[TD]

Use the property of logs that log(xy) = log(x) + log(y) with the double angle formula of the sine.

 

[Gib Z]

I=\frac{1}{4} \int^{\pi/2}_0 \log_e (\frac{1}{2}\sin 2x) dx...I don't get it...

 

[Gib Z]

Wait up I think I am getting it, Ill post it in 5 mins

 

[Gib Z]

Nope lost again...

 

[Hootenanny]

Umm Hootenannys one took a while, I Basically did t=tan (x/2) and made it a rational function so on so forth.

I got \frac{2}{\sqrt{15}} \arctan \frac{4 \tan (x/2) +1}{\sqrt{15}} + C which seems like a hairy answer, I don't think I am right. I can't be stuffed to differentiate it to check lol.

Absolutely spot on Gib, it can't be simplified any further

 

[morphism]

Hehe the first one is easier than I thought. I first tried u = arctan x + arccot x, but when I tried to find du I realized the derivative was 0, so I finally noticed it was a constant function. I tried x=1, the constant function is equal to pi/2, so the solution is \frac{\pi^2}{4}!

Yes, that's right, but there's a solution that'st truly one ling long. Yours involves differentiating things and evaluating it at a point, so it's longer than one line, especially if you don't know the derivatives of those things. The solution I have in mind is: Notice that arctan(x) and arccot(x) over [0, pi/2] correspond to the two accute angles in a right-angled triangle. So they add up to pi/2.

I couldn't do the 2nd one :(

How about setting x=tan(t)?

Third one I did u= log (2), du/dx=1/x which is in there so all good.
So I integrated it and ended up with \frac{\log^2 (x+1)}{2}.

I'm not sure I follow. Do you mean you set u=log(x)? That doesn't help...

Last one, I split it up into 2 integrals, from 0 to pi/2, and pi/2 to pi.
Using symmetry arguments and the such, I got that it 2I= \int^{\pi/2}_0 \log(\sin x) dx = \int^{\pi/2}_0 \log(\cos x) dx but I am stuck from there.

That's good. Now use these to get I. (sin(x) = 2sin(x/2)cos(x/2) and log(xy) = log(x)+log(y).)

 

[Gib Z]

The third one I sed u = log(x+1), and that helps because du/dx = 1/x, which is there as well. So My solution for it was \frac{\ln 2)^2}{2}.

Last one, that's what i did and i got this!

I=\frac{1}{4} \int^{\pi/2}_0 \log_e (\frac{1}{2}\sin 2x) dx

Can't do that one!

 

[Gib Z]

For the one you said x=tan t, I did it but I can finish off \int \ln (\tan t +1) dt

 

[morphism]

The third one I sed u = log(x+1), and that helps because du/dx = 1/x, which is there as well. So My solution for it was \frac{\ln 2)^2}{2}.

But du/dx isn't 1/x! It's 1/(x+1).

Last one, that's what i did and i got this!

I=\frac{1}{4} \int^{\pi/2}_0 \log_e (\frac{1}{2}\sin 2x) dx

Can't do that one!

OK, let's see.

I = \int_0^{\pi} \log(2) + \log(\sin \frac{x}{2}) + \log(\cos \frac{x}{2} ) \; dx

From your observations, we can see that this is just I = pi log(2) + I + I.

 

[Gib Z]

MY GOD KILL ME, I really should have seen those >.< Staring me in the face..

Okie well its getting late here in Australia, Ill try again with the remaining 2 tomorrow.

 

[morphism]

For the one you said x=tan t, I did it but I can finish off \int \ln (\tan t +1) dt

\ln (1 + \tan t) = \ln(\sin t + \cos t) - \ln(\cos t) = \ln \left( \sqrt2\cos(t - \frac{\pi}{4}) \right) - \ln(\cos t)

Good night!

 

[Gib Z]

Ok I am back But for your hint on the x=tan t one, am I just awfully stupid or is that still quite hard to integrate..

 

[morphism]

Here's the next step:

\ln(\sqrt 2) + \ln \left( \cos \left( t - \frac{\pi}{4} \right) \right) - \ln(\cos t)

Now use the limits of integration...

 

[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.