How Good Am I at Integrals?

  • Thread starter Gib Z
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In summary, the individual tried to do an integrative problem involving tan x, but was not able to get the answer correct. They substituted for dx and got an incorrect answer.
  • #246
The integral [tex]\int_0^1\frac{ln(1+x)}{x}\cdot dx[/tex] can be evaluated by considering the series expansion of the numerator, i.e. [tex]ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\cdot \frac{x^k}{k}[/tex]. Putting this into the integral gives after dividing it by [tex]x[/tex], switching the integral and the sum (permitted due to converging series): [tex]\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}x^{k-1}dx[/tex]. This is after evaluation: [tex]\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}[/tex]. This can be split up into the following two series [tex]\left(1+\frac{1}{3^2}+\frac{1}{5^2}+...\right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\right)[/tex].
These are respectively equal to [tex]\frac{\pi^2}{8}[/tex] and [tex]\frac{\pi^2}{24}[/tex]. The result is therefore [tex]I=\frac{\pi^2}{12}[/tex]. The series are in my personal formula map :-), I think they were comming from a Fourier series of some kind, I don't exactly remember.

It was a nice exercise.

The other integral [tex]\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{sin(p\pi)}[/tex] seems to be wrong to me. Isn't the power supposed to be [tex]p-1[\tex]? Then it can be transformed into a beta function and via complex contour integration solved for the result shown. Using the power given as it is, will result in something different. This can be found in more advanced textbooks :-) All this under the assumption that I'm not mistaken, it's been a while since I have done these kind of things. If I find the time I will try to work it out in detail.
 
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  • #247
Count Iblis said:
[tex]\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{\sin\left(\pi p\right)}[/tex] :smile:

If p is an integer, that is obviously false as sin(p*pi)=0.
 
  • #248
The integral converges for [tex]0<p<1[/tex]. Also, to address the point raised by coomast, changing p to p-1 would introduce a minus sign. Sometimes you see the numerator given as [tex]x^{p-1}[/tex]. If we change this to [tex]x^{p}[/tex] we get a minus sign and if we change p to minus p we get another minus sign. To check the sign, note that the integral is positive if it converges and so is the answer.
 
  • #249
my favorite way to do odd powers of sec is by parts. it reduces the integral's power by 2 in each step. until finally you need to know just the integral of sec. (i hate double angle formulas in integrals as they make them seem complicated and tedious).
 
  • #250
I agree with Mathwonk. I find integration by parts to be easier it comes to sec with odd powers. I'm pretty bad at remembering the trig tricks, so in the end, i'll goof it up. I like the safe route.
 
  • #251
coomast said:
The integral [tex]\int_0^1\frac{ln(1+x)}{x}\cdot dx[/tex] can be evaluated by considering the series expansion of the numerator, i.e. [tex]ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\cdot \frac{x^k}{k}[/tex]. Putting this into the integral gives after dividing it by [tex]x[/tex], switching the integral and the sum (permitted due to converging series): [tex]\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}x^{k-1}dx[/tex]. This is after evaluation: [tex]\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}[/tex]. This can be split up into the following two series [tex]\left(1+\frac{1}{3^2}+\frac{1}{5^2}+...\right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\right)[/tex].
These are respectively equal to [tex]\frac{\pi^2}{8}[/tex] and [tex]\frac{\pi^2}{24}[/tex]. The result is therefore [tex]I=\frac{\pi^2}{12}[/tex]. The series are in my personal formula map :-), I think they were comming from a Fourier series of some kind, I don't exactly remember.

It was a nice exercise.

That is the correct evaluation of the Integral, although (don't hate me for this, I'm reading Courant at the moment) the proof can be made more rigorous by first stating that the Series expansion for the logarithm is valid for all points within the interval of integration, justifying taking the series out of the integral or just integrating the sum term by term from the inside, and rearranging the resulting series to be shown to be [itex]\frac{\zeta (2)}{2}[/itex] which indeed gives [itex]\pi^2/6[/itex] although the value of zeta 2 is much more well known.
 
  • #252
I'm not hating you for putting attention on the important details Gib Z. I always check them, but don't post everything explicitely.

Edit: Mmmm, misread, I suppose you meant that you had a slightly different method. This is always welcome :-)

The other integral from Count Iblis can indeed be solved by starting from the Beta function. I'm not going to do the entire thing here, but it goes something like this. The definition of the beta function is [tex]\int_{0}^{1}t^{m-1}(1-t)^{n-1}dt[/tex]. Using the following substitution [tex]m=1-p[/tex] and [tex]n=p[/tex] gives [tex]\Gamma(p)\Gamma(1-p)[/tex], which can be shown to be [tex]\frac{\pi}{sin(p\pi)}[/tex]. This is done by contour integration. (Details can be found in Spiegel).

The integral of the beta function can be transformed using the substitution [tex]u=\frac{t}{1-t}[/tex] which transforms it into the Count Iblis one.

Is this OK for a proof? It uses a different function as a workaround and I can imagine that simpler ways exist.

Now the mistake I made with the power was nothing more than using [tex]p=1-y[/tex], which is also between 0 and 1. It transforms the integral into something that is more commonly found in the literature. The results are however indeed the same as allready pointed out by Count Iblis. [tex]sin((1-p)\pi)=sin(p\pi)[/tex].

The following can also be shown to hold: [tex]\int_{0}^{\infty}\frac{x^{-p}}{x-1}dx=\pi\cdot cot(p\pi)[/tex] Also for 0<p<1.

Finally the following one can be entirely solved as a "real" one, for the ones who have some spare time :-)

[tex]\int \frac{dx}{\sqrt{1+sin(x)}}[/tex]

Hope you enjoy it. I will not be able to post for two weeks, too much work at my job :-(
 
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  • #253
coomast said:
I'm not hating you for putting attention on the important details Gib Z. I always check them, but don't post everything explicitely.

Edit: Mmmm, misread, I suppose you meant that you had a slightly different method. This is always welcome :-)

The other integral from Count Iblis can indeed be solved by starting from the Beta function. I'm not going to do the entire thing here, but it goes something like this. The definition of the beta function is [tex]\int_{0}^{1}t^{m-1}(1-t)^{n-1}dt[/tex]. Using the following substitution [tex]m=1-p[/tex] and [tex]n=p[/tex] gives [tex]\Gamma(p)\Gamma(1-p)[/tex], which can be shown to be [tex]\frac{\pi}{sin(p\pi)}[/tex]. This is done by contour integration. (Details can be found in Spiegel).

The integral of the beta function can be transformed using the substitution [tex]u=\frac{t}{1-t}[/tex] which transforms it into the Count Iblis one.

Is this OK for a proof? It uses a different function as a workaround and I can imagine that simpler ways exist.

Now the mistake I made with the power was nothing more than using [tex]p=1-y[/tex], which is also between 0 and 1. It transforms the integral into something that is more commonly found in the literature. The results are however indeed the same as allready pointed out by Count Iblis. [tex]sin((1-p)\pi)=sin(p\pi)[/tex].

The following can also be shown to hold: [tex]\int_{0}^{\infty}\frac{x^{-p}}{x-1}dx=\pi\cdot cot(p\pi)[/tex] Also for 0<p<1.

Finally the following one can be entirely solved as a "real" one, for the ones who have some spare time :-)

[tex]\int \frac{dx}{\sqrt{1+sin(x)}}[/tex]

Hope you enjoy it. I will not be able to post for two weeks, too much work at my job :-(

[tex]\int \frac{dx}{\sqrt{1+sin(x)}}[/tex]=[tex]\int \frac{dx}{\sqrt{sin^{2}(x/2)+cos^{2}(x/2)+2sin(x/2)cos(x/2)}}[/tex]=[tex]\int \frac{dx}{\sqrt{(cos(x/2)+sin(x/2))^{2}}}[/tex]=[tex]\int \frac{dx}{\(cos(x/2)+sin(x/2))}[/tex]

now let us take the substitution x/2=t, dx=2dt, and we get

2[tex]\int \frac{dt}{\(cos(t)+sin(t)}[/tex]
now again let's take the substitution tg(x/2)=u, dt=2du/(1+u^2), and sin(t)=2t/(1+t^2) and cos(t)=(1-t^2)/(1+t^2), and after some elementary transformations we get:
4[tex]\int \frac{du}{\(2u+1-u^{2}}[/tex]

After we use parcial fractions somewhere on the way we finally get
:

-(sqrt2)* ln I(tg(x/4)-(1-sqrt2))/tg(x/4)-(1+sqrt2)I
 
  • #254
coomast said:
I'm not hating you for putting attention on the important details Gib Z. I always check them, but don't post everything explicitely.

Edit: Mmmm, misread, I suppose you meant that you had a slightly different method. This is always welcome :-)

The other integral from Count Iblis can indeed be solved by starting from the Beta function. I'm not going to do the entire thing here, but it goes something like this. The definition of the beta function is [tex]\int_{0}^{1}t^{m-1}(1-t)^{n-1}dt[/tex]. Using the following substitution [tex]m=1-p[/tex] and [tex]n=p[/tex] gives [tex]\Gamma(p)\Gamma(1-p)[/tex], which can be shown to be [tex]\frac{\pi}{sin(p\pi)}[/tex]. This is done by contour integration. (Details can be found in Spiegel).

The integral of the beta function can be transformed using the substitution [tex]u=\frac{t}{1-t}[/tex] which transforms it into the Count Iblis one.

Is this OK for a proof? It uses a different function as a workaround and I can imagine that simpler ways exist.

Now the mistake I made with the power was nothing more than using [tex]p=1-y[/tex], which is also between 0 and 1. It transforms the integral into something that is more commonly found in the literature. The results are however indeed the same as allready pointed out by Count Iblis. [tex]sin((1-p)\pi)=sin(p\pi)[/tex].

The following can also be shown to hold: [tex]\int_{0}^{\infty}\frac{x^{-p}}{x-1}dx=\pi\cdot cot(p\pi)[/tex] Also for 0<p<1.

Finally the following one can be entirely solved as a "real" one, for the ones who have some spare time :-)

[tex]\int \frac{dx}{\sqrt{1+sin(x)}}[/tex]

Hope you enjoy it. I will not be able to post for two weeks, too much work at my job :-(

[tex]\int \frac{dx}{\sqrt{1+sin(x)}}[/tex]=[tex]\int \frac{dx}{\sqrt{sin^{2}(x/2)+cos^{2}(x/2)+2sin(x/2)cos(x/2)}}[/tex]=[tex]\int \frac{dx}{\sqrt{(cos(x/2)+sin(x/2))^{2}}}[/tex]=[tex]\int \frac{dx}{\(cos(x/2)+sin(x/2))}[/tex]

now let us take the substitution x/2=t, dx=2dt, and we get

2[tex]\int \frac{dt}{\(cos(t)+sin(t)}[/tex]
now again let's take the substitution tg(x/2)=u, dt=2du/(1+u^2), and sin(t)=2t/(1+t^2) and cos(t)=(1-t^2)/(1+t^2), and after some elementary transformations we get:
4[tex]\int \frac{du}{\(2u+1-u^{2}}[/tex]

After we use parcial fractions somewhere on the way we finally get
:

-(sqrt2)* ln I(tg(x/4)-(1-sqrt2))/tg(x/4)-(1+sqrt2)I


I hope my answer is right. NOw i am going to post two problems:

1. [tex]\int[x]dx[/tex], where x>=0
 
  • #255
And the other problem is:

2. Let f:(0,1]-->R (reals) be a function with continuous derivatives, so that f(0)=0

Show that for every n form naturals we have:

[tex]\int_0^1\{f'(1-x^{n})x^{2n-1}}\ dx[/tex] =1/n [tex]\int_0^1\{f(x)}\ dx[/tex]
 
  • #256
Define [x] please =]? I've seen that to mean either the ceiling or floor functions.
 
  • #257
Gib Z said:
Define [x] please =]? I've seen that to mean either the ceiling or floor functions.

Well, [x] is supposed to be the floor function. I did not say this, because usually when we wrote the function like this we assumed to be the floor function. Hwever, my bad, i think i should have defined it here.
 
  • #258
I should be doing other stuff, but I just couldn't resist...

First the solution given by sutupidmath. It is exactly the same method that I used. However I don't have the minus sign at the final solution. I will check it later on. You can imagine that the first step the crucial one is. It can be painfull, if one doesn't see it :-)

The second question from sutupidmath can be proven by using the following substitution [tex]1-x^n=y[/tex] in the left hand side integral.
This gives now [tex]\frac{1}{n}\int_{0}^{1}f'(y)\cdot (1-y)\cdot dy[/tex]. This can be rewritten using partial integration to give:
[tex]\frac{1}{n}\int_{0}^{1}f'(y)\cdot dy-\frac{1}{n}\int_{0}^{1}f'(y)\cdot y \cdot dy=\frac{1}{n}(f(1)-f(0))-\frac{1}{n}\left(\left[yf(y)\right]_{0}^{1}-\int_{0}^{1}f(y)\cdot dy\right)[/tex]

This is then, because f(0)=0:

[tex]\frac{1}{n}f(1)-\frac{1}{n}f(1)+\frac{1}{n}\int_{0}^{1}f(y)\cdot dy=\frac{1}{n}\int_{0}^{1}f(y)\cdot dy[/tex]

Because y is a "dummy" integral variable, it can be replaced by x, giving the result.
I tried it on a simple example and it works, very nice formula.
 
  • #259
[tex] \int_{0}^{\pi/2} ab \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt [/tex]
 
  • #260
sutupidmath said:
1. [tex]\int[x]dx[/tex], where x>=0

If you guys want i could throw some hints?
 
  • #261
tronter said:
[tex] \int_{0}^{\pi/2} ab \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt [/tex]
i thinki got it. I am going to post it very briefly because my friend is calling me up
 
  • #262
[tex] \int_{0}^{\pi/2} \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt [/tex]=ab[tex] \int_{0}^{\pi/2} \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt [/tex]=ab/2 [tex] \int_{0}^{\pi/2} 2( \sin t \cos t) \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt [/tex]=ab/2 [tex] \int_{0}^{\pi/2} ( \sin 2t ) \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt [/tex]=ab/2 [tex] \int_{0}^{\pi/2} ( \sin 2t ) \sqrt{a^{2} \((1-cos 2t)/2 + b^{2} \((1+cos2t)/2} \ dt [/tex]=
then after some other transformaions i come to this point
ab/sqrt2 [tex] \int_{0}^{\pi/2} ( \(sin 2t)/2 ) \sqrt{cos2t(b^{2}-a^{2})+a^{2}+b^{2}} \ dt [/tex]=
then i take this substitution

(b^2-a^2)cos2t+a^2+b^2=u, where sin2t/2 =du/(b^2-a^2)

and so on, i appologize for not having time to go to the end, i did it on my paper, however no time to type all of it

edit:

i"ll keep going anyway
ab/(a^2-b^2)sqrt2 [tex] \int_{2b^{2}}^{\2a^{2}} ({ \sqrt u} \ du [/tex]=

SO now it gets pretty easy to integrate.
 
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  • #263
the upper bound here should be 2a^2
 
  • #264
I'm not too sure what exactly you did, but there is an easier way :(

Rewrite the integral as such (and remember [itex]\sin 2t = 2\sin t \cos t[/itex] because I use that a few times):

[tex]\frac{ab}{2} \int^{\pi/2}_0 \sin (2t) \sqrt{ a^2 \sin^2 t + b^2 \cos^2 t} dt[/tex].

Let [itex]u= a^2 \sin^2 t + b^2 \cos^2 t[/itex], then [tex] dt = \frac{du}{\sin (2t) \cdot (a^2-b^2)}[/tex] so that ( o yea I'll change bounds now as well):

[tex] \frac{ab}{2} \int^{a^2}_{b^2} \sin (2t) \sqrt{u} \frac{du}{\sin (2t) \cdot (a^2-b^2)} = \frac{ab}{2(a^2-b^2)} \int^{a^2}_{b^2} \sqrt{u} du = \frac{ab}{2(a^2-b^2)} \cdot \frac{2(a^3-b^3)}{3} = \frac{(ab)(a^2+ab+b^2)}{3(a+b)} [/tex]

Note: It seems we get different answers :(
 
  • #265
sutupidmath said:
If you guys want i could throw some hints?

Yes I wouldn't mind a hint on that one, so far I can only get an anti derivative valid for a periodic interval of 2pi using Fourier series :(
 
  • #266
Gib Z said:
I'm not too sure what exactly you did, but there is an easier way :(

Rewrite the integral as such (and remember [itex]\sin 2t = 2\sin t \cos t[/itex] because I use that a few times):

[tex]\frac{ab}{2} \int^{\pi/2}_0 \sin (2t) \sqrt{ a^2 \sin^2 t + b^2 \cos^2 t} dt[/tex].

Let [itex]u= a^2 \sin^2 t + b^2 \cos^2 t[/itex], then [tex] dt = \frac{du}{\sin (2t) \cdot (a^2-b^2)}[/tex] so that ( o yea I'll change bounds now as well):

[tex] \frac{ab}{2} \int^{a^2}_{b^2} \sin (2t) \sqrt{u} \frac{du}{\sin (2t) \cdot (a^2-b^2)} = \frac{ab}{2(a^2-b^2)} \int^{a^2}_{b^2} \sqrt{u} du = \frac{ab}{2(a^2-b^2)} \cdot \frac{2(a^3-b^3)}{3} = \frac{(ab)(a^2+ab+b^2)}{3(a+b)} [/tex]

Note: It seems we get different answers :(


well our answers do not differ that much, at the very end i got

[tex]\frac{4(ab)(a^2+ab+b^2)}{3(a+b)} [/tex]
Either i have added something, or you have omitted sth. Anyway, it is quite likely that i have not simplified, or canceled out sth on the way, because i did it in less than 3 min, of course when i did it on my paper, because it took me 30 min just to type it, i am terrible at latex.
However it may be, i am not going to bother and look at it again, it is a technical mistake.
 
  • #267
Gib Z said:
Yes I wouldn't mind a hint on that one, so far I can only get an anti derivative valid for a periodic interval of 2pi using Fourier series :(

well on this one let [x]=t, and let x be from the interval (t, t+1)

now evaluate the integral F(x)=integ from 0 to x of [t] dt
 
  • #268
ansrivas said:
[tex] I=\int \frac{\sin \theta - \cos \theta}{(\sin \theta + \cos \theta)\sqrt{\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta}} d \theta[/tex]
[tex] = \int \frac{\sin^2 \theta - \cos^2 \theta}{(1 + 2\sin \theta \cos \theta)\sqrt{\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta}} d \theta[/tex]

Now let [tex] u=\sin \theta \cos \theta [/tex]
[tex]\frac{du}{d\theta}=\cos^2\theta-\sin^2\theta[/tex]

This gives

[tex]I=\int \frac{-1}{(1+2u)\sqrt{u+u^2}}\,du [/tex]

Now let [tex]l=\sqrt{u+u^2}[/tex]
[tex]\frac{dl}{du}=\frac{1+2u}{2\sqrt{u+u^2}}[/tex]

This gives
[tex]I=\int \frac{-2}{4l^2+1}dl [/tex]
i did it a different way, and with help from my teacher for the last few steps. thanks Sidd, i learned a lot from your so called "simple" problem :p

http://alt1.mathlinks.ro/Forum/latexrender/pictures/8/4/a/84acab97b0086fb490dcb2c9860e9d8c2026656a.gif [Broken]
 
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  • #269
the title of this thread always tempts me to answer "you stink!". but i won't do that.

the other temptation is to respond "you are great!" but that would be spoiling.

so ill suggest: you are good but not great, like most wines.
 
  • #270
Was that aimed at me or rocophysics? I know I stink :( Whats a wine?
 
  • #271
Gib Z said:
Was that aimed at me or rocophysics? I know I stink :( Whats a wine?
It was aimed at you, I've already spoiled :p
 
  • #272
How good are YOU ?

Here are some of my favourite integrals, ranging from beginners to just no.
Some of these have been posted before, most have not. Have fun.

Have fun and gl

Easier problems at the top, harder problems at the bottom. I really love all of these

Easy

[tex]I \, = \, \int {\frac{x}{e^x}} dx [/tex]
[tex]I \, = \, \int_0^{\frac{\pi }{3n}} \tan \left( {nx} \right) dx [/tex]
[tex]I \, = \, \int \frac{x^2+1}{x(x^2+3)} dx [/tex]
[tex]I \, = \, 1+\int_{229}^{1234} (\sin(x)+\cos(x))^2+(\cos(x)-\sin(x))^2 \, dx[/tex]
[tex]I \, = \, \int {\frac{1}{x\ln x}} dx [/tex]
[tex]I \, = \, \int {\frac{{{e^x} + 1}}{{{e^x} - 1}}} dx [/tex]
[tex]I \, = \, \int \frac{\sin(x)}{\sin(x)-1} dx [/tex]
[tex]I \, = \, \int {2^{\ln(x)}} dx [/tex]
[tex]I \, = \, \int {{e^{x + {e^x}}}} dx [/tex]
[tex]I \, = \, \int_0^4 {\frac{\ln \left( x \right)}{\sqrt x}} dx [/tex]
[tex]I \, = \, \int {{e^{\sqrt x }}} dx [/tex]
[tex]I \, = \, \int \frac{1}{x^7-x} dx [/tex]
[tex]I \, = \, \int\limits_0^4 {\frac{1}{{1 + \sqrt x }}} dx [/tex]
[tex] I \, = \, \int \frac{\cos \left( x \right) - \sin \left( x \right)}{\sin \left( x \right) + \cos \left( x \right)} \, dx [/tex]
[tex]I \, = \, \int \frac{\sin(x)\cos(x)}{\cos(x)^4-\sin(x)^4} \, dx[/tex]
[tex] I \, = \, \int \frac{\sqrt{\sqrt{\ln(x)}+1}}{x} dx [/tex]
[tex] I \, = \, \int \frac{1}{1+\sin\left( \frac{\pi}{6} \right)^x} dx [/tex]
[tex]I \, = \, \int {\frac{1}{x \ln{{\left( x \right)}^n}}} dx [/tex]
[tex]I \, = \, \int {\frac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx [/tex]
[tex]I \, = \, \int \frac{2010}{x(1+x^{2010})} dx [/tex]

Find the area between the function [itex]f(x)[/itex] and the x-axis when [itex]f(x)=\sqrt{a-\sqrt{x}}[/itex]

Show that [itex] \int \left( {x + 3} \right) \left( {x - 1} \right)^5 dx \: [/itex] equals [itex] \: \frac{1}{21}\left( {3x + 11} \right){{\left( {x - 1} \right)}^6} + C [/itex]

Medium

[tex] I \, = \, \int \sin(\ln(x)) + \cos(\ln(x)) dx [/tex]
[tex] I \, = \, \int {\left( {1 + 2{x^2}} \right){e^{{x^2}}}} \, dx [/tex]
[tex] I \, = \, \int \frac{\ln(x)-1}{\ln(x)^2} \, dx [/tex]
[tex] I \, = \, \int_{\pi}^{3\pi} \sin(x)\ln(x) - \frac{\cos(x)}{x} \, dx [/tex]
[tex] I \, = \, \int \frac{1}{\ln(x)}+\ln(\ln(x)) \, dx [/tex]
[tex] I \, = \, \int (x+3) \sqrt{e^x \cdot x} dx [/tex]
[tex] I \, = \, \int \arccos(x) + \arcsin(x) dx [/tex]
[tex] I \, = \, \int \frac{1}{\cos(x)} dx [/tex]
[tex] I \, = \, \int \frac{4(\ln x)^2+1}{(\ln x)^{\frac 32}}\ dx [/tex]
[tex] I \, = \, \int \frac{1}{\sqrt[3]{x}+x}\, dx [/tex]
[tex] I \, = \, \int \frac{\sin(x)+\sin(3x)}{\cos(3x)+\cos(x)} \, dx [/tex]
[tex] I \, = \, \int \frac{1}{\sqrt{x+x\sqrt{x}}} \, dx [/tex]
[tex] I \, = \, \int \sin(101x)\cdot\sin(99x) \, dx [/tex]
[tex] I \, = \, \int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} \, dx [/tex]
[tex] I \, = \, \int_0^1 \frac{x}{1+x}\sqrt{1-x^2}\ dx [/tex]
[tex] I \, = \, \int \frac{1}{x^{a+1}+x} \, dx \; a>0[/tex]
[tex] I \, = \, \int \frac{2^x \cdot 3^x}{9^x- 4^x} \, dx [/tex]
[tex] I \, = \, \int \frac{1 + 2x^2}{x^5 \left( 1 + x^2 \right)^3} \, dx [/tex]
[tex] I \, = \, \int \frac{1}{\sqrt{5x+3}-\sqrt{5x-2}} \, dx [/tex]
[tex] I \, = \, \int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^2+1}}[/tex]
[tex] I \, = \, \int_1^e \frac{\ln x-1}{x^2-(\ln x)^2}dx[/tex]
[tex] I \, = \, \int_{1}^{\sqrt{3}} x^{2x^2+1}+\ln(x^{2x^{2x^2+1}}) dx[/tex]

The integral [itex]\; \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2} \, dx } [/itex]
can be written as [itex]\; \left( \frac{a}{b} \right)^b [/itex] where a and b are integers. Find [itex]\sqrt{a^b+b^a+1}[/itex]

[tex] I \, = \, \int \frac{\left( x + 2 \right)^5}{\left( x + 7 \right)^2} \, dx [/tex]
[tex] I \, = \, \int \frac{\sinh(x)+\cosh(x)}{\cosh(x)-\sinh(x)} \, dx [/tex]
[tex]I \, = \, \int\limits_0^{\pi /2} {\sqrt {1 - 2\sin \left( {2x} \right) + 3\cos {{\left( x \right)}^2}} dx} [/tex]
[tex]I \, = \, \int_{0}^{\pi/2} \ln(sec(x))[/tex]
[tex]I \, = \, \int_{0}^{a} \frac{1}{\sqrt{x^2+a^2}} dx [/tex]
[tex] I \, = \, \int \sqrt{ \sqrt{x+2\sqrt{2x+4}} + \sqrt{x-2\sqrt{2x+4}} } dx[/tex]
[tex] I \, = \, \int \frac{ \left( 1 + x^2 \right) } { \left( 1 - x^2 \right) \sqrt{1 + x^4} } \, dx [/tex]
[tex] I \, = \, \int \frac{ \sin(x) + \cos(x) }{e^x + 3 \cos(x) } \, dx [/tex]
[tex] I \, = \, \int \sqrt{ \frac{k + x}{x} } \, dx [/tex]
[tex] I \, = \, \int_{0}^{\infty} x^n \cdot e^{-x} \, dx [/tex]
[tex] I \, = \, \int {\frac{x}{1+\cos(x)}} \, dx [/tex]
[tex] I \, = \, \int {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \, dx [/tex]
[tex] I \, = \, \int\limits_{ - \pi }^\pi {\sqrt {1 + \cos \left( x \right)} } \, dx [/tex]
[tex] I \, = \, \int e^{x/2} \cdot \left( \frac{2 - \sin \left( x \right) }{1 - \cos \left( x\right) } \right) \, dx [/tex]
[tex] I \, = \, \int_{0}^{\infty} \frac{\{1-(x-1)e^x\}\ln(x)}{(1+e^x)^2} dx [/tex]
[tex] I \, = \, \int\limits_0^{\ln \left( 2 \right)} {\sqrt {\frac{{{e^x} + 1}}{{{e^x} - 1}}} } \, dx [/tex]
[tex] I \, = \, \int_{0}^{\infty} \frac{1}{ \left( x + \sqrt{1+x^2} \right)^{\phi}} \, dx \qquad \phi=\frac{1+\sqrt{5}}{2}[/tex]
[tex] I \, = \, \int_0^{\infty} \frac{1}{(x^2+2)^3} \, dx [/tex]
[tex]I \, = \, \int \sqrt{\frac{1}{\,\sin(x)+1}\,}\, dx[/tex]
[tex] I \, = \, \int {\sqrt[3] {\tan \left( x \right)} } \, dx [/tex]
[tex] I \, = \, \int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}}\, dx \, = \, \frac{14!}{2 \cdot 5^{15} \cdot 49^2 \cdot 16! }\, dx [/tex]
[tex] I \, = \, \int \left( \frac{\arctan(x)}{x-\arctan(x)} \right)^2 \, dx [/tex]

"Fun"

[tex]I \, = \, \int_{1}^{\infty} \ln\left( \sqrt[X]{x} \right)^{2011} dx [/tex]
[tex]I \, = \, \int_{0}^{2\pi} \frac{x \sin(x)}{1+\sin(x)^2} dx [/tex]
[tex]I \, = \, \int_{0}^{2\pi} {\frac{1}{1+e^{\cos(x)}}} dx [/tex]
[tex]I \, = \, \int_0^{\frac {\pi}{2}} \frac{(\sin x)^{\cos x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx [/tex]
[tex]I \, = \, \int_2^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}} dx [/tex]
[tex]I \, = \, \int_{-1}^{1} \arctan(e^x) dx [/tex]
[tex]I \, = \, \int_{0}^{1} \ln(x)\cdot\ln(1-x) dx [/tex]
[tex]I \, = \, \int_{-\pi/2}^{\pi/2} \frac{1}{1+2009^x} \cdot \frac{\sin(2010x)}{\sin(2010x)+\cos(2010x)} dx [/tex]
[tex]I \, = \, \int \frac{1}{(x^2+1)^k} dx [/tex]
[tex]I \, = \, \int_{0}^{\infty} \frac{x}{e^x-1} dx [/tex]
[tex]I \, = \, \int_{0}^{a} \frac{1}{x+\sqrt{a^2-x^2}} dx \quad a>0[/tex]
[tex]I \, = \, \int_{0}^{\infty} \frac{\ln{2x}}{x^2+9} dx [/tex]
[tex] I \, = \, \int_{0}^{\pi/2} \ln(1-a \cos(x)) \, dx [/tex]
[tex]I \, = \, \int_{-\infty}^{\infty} \sin(x^2)+\cos(x^2) dx [/tex]
[tex]I \, = \, \int_{-\infty}^{\infty} \frac{x^a+x^b}{\ln(x)} dx \qquad (a,b)\forall >-1[/tex]
[tex]I \, = \, \int_0^1 \frac{\ln(1+x^2)}{1+x^2} dx [/tex]
[tex]I \, = \, \int_{0}^{\infty}\frac{1}{x^n+1} dx \quad n>1[/tex]
[tex]I \, = \, \int_{0}^{\pi} \frac{1-\cos{(n\cdot x)}}{1-\cos(x)} dx \quad n\in\mathbb{N^+}[/tex]
[tex]I \, = \, \int_{0}^{1} \ln(x) \cdot e^{-x} \cdot (1-x) dx [/tex]
[tex]I \, = \, \int_{0}^{\infty} \ln\left( \frac{e^x+1}{e^x-1}\right) dx[/tex]
[tex]I \, = \, \int_{0}^{\pi} \ln \left( 1 - 2\alpha \cos(x) + \alpha^2 \right) dx[/tex]
[tex]I \, = \, \int_0^{\infty} \ln (1+e^{-ax} ) dx \quad a>0[/tex]
[tex]I \, = \, \int_0^{\infty} 1 - x\sin\left( \frac{1}{x} \right) dx[/tex]
[tex]I \, = \, \int_0^{\infty} \ln (1+e^{-ax} ) dx \quad a>0[/tex]
[tex] I = \int_{0}^{\infty} \frac{{{e^{ - x}}\left( {1 - {e^{ - 6x}}} \right)}}{{x\left( {1 + {e^{ - 2x}} + {e^{ - 4x}} + {e^{ - 6x}} + {e^{ - 8x}}} \right)}}dx[/tex]
[tex] I = \int\limits_0^1 {\frac{{\sin \left( {p\ln x} \right) \cdot \cos \left( {q\ln x} \right)}}{{\ln x}}}dx[/tex]
[tex]I \, = \, \int_{a}^{\infty} \frac{1}{x^{n+1} \cdot \sqrt{x^2-a^2}} \, dx n \in \mathbb{N}^{+} \; , \; a>0[/tex]
[tex]I \, = \, \int_{a}^{\infty} \frac{x^{-p}}{1+x} \, dx[/tex]
[tex]I \, = \, \int_{ - \infty }^{\infty} \frac{x\sin x}{1 + x^2}dx [/tex]
[tex]I \, = \, \int_0^{\infty} \sin(x)\cdot\arctan\left( \frac{1}{x} \right) dx[/tex]
[tex]I \, = \, \int_{0 }^{\infty} \frac{1}{x} \cdot \sin(\tan(x)) \, dx [/tex]

Have fun and gl.
 
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<h2>1. How do I know if I am good at integrals?</h2><p>Your proficiency in integrals can be determined by your ability to solve a variety of integral problems accurately and efficiently. Additionally, your understanding of the fundamental concepts and techniques of integration can also indicate your level of skill.</p><h2>2. What are the most important skills for mastering integrals?</h2><p>Some of the key skills for becoming proficient in integrals include a strong foundation in algebra and trigonometry, a good understanding of the properties of functions, and the ability to apply various integration techniques such as substitution, integration by parts, and partial fractions.</p><h2>3. How much practice is needed to become good at integrals?</h2><p>Becoming proficient in integrals requires consistent practice and exposure to a variety of integral problems. The amount of practice needed may vary for each individual, but it is important to practice regularly and challenge yourself with increasingly difficult problems.</p><h2>4. Are there any tips or tricks for solving integrals more easily?</h2><p>Yes, there are a few tips and tricks that can help make solving integrals easier. These include recognizing patterns in integrals, using substitution to simplify complex integrals, and breaking down integrals into smaller, more manageable parts.</p><h2>5. How can I improve my speed and accuracy in solving integrals?</h2><p>One way to improve your speed and accuracy in solving integrals is to practice regularly and familiarize yourself with common integration techniques. Additionally, it can be helpful to work on your mental math skills and learn shortcuts for solving certain types of integrals.</p>

1. How do I know if I am good at integrals?

Your proficiency in integrals can be determined by your ability to solve a variety of integral problems accurately and efficiently. Additionally, your understanding of the fundamental concepts and techniques of integration can also indicate your level of skill.

2. What are the most important skills for mastering integrals?

Some of the key skills for becoming proficient in integrals include a strong foundation in algebra and trigonometry, a good understanding of the properties of functions, and the ability to apply various integration techniques such as substitution, integration by parts, and partial fractions.

3. How much practice is needed to become good at integrals?

Becoming proficient in integrals requires consistent practice and exposure to a variety of integral problems. The amount of practice needed may vary for each individual, but it is important to practice regularly and challenge yourself with increasingly difficult problems.

4. Are there any tips or tricks for solving integrals more easily?

Yes, there are a few tips and tricks that can help make solving integrals easier. These include recognizing patterns in integrals, using substitution to simplify complex integrals, and breaking down integrals into smaller, more manageable parts.

5. How can I improve my speed and accuracy in solving integrals?

One way to improve your speed and accuracy in solving integrals is to practice regularly and familiarize yourself with common integration techniques. Additionally, it can be helpful to work on your mental math skills and learn shortcuts for solving certain types of integrals.

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