Are You Ready to Challenge Your Integral Solving Skills?

[coomast]

The integral $$\int_0^1\frac{ln(1+x)}{x}\cdot dx$$ can be evaluated by considering the series expansion of the numerator, i.e. $$ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\cdot \frac{x^k}{k}$$. Putting this into the integral gives after dividing it by $$x$$, switching the integral and the sum (permitted due to converging series): $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}x^{k-1}dx$$. This is after evaluation: $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}$$. This can be split up into the following two series $$\left(1+\frac{1}{3^2}+\frac{1}{5^2}+...\right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\right)$$.
These are respectively equal to $$\frac{\pi^2}{8}$$ and $$\frac{\pi^2}{24}$$. The result is therefore $$I=\frac{\pi^2}{12}$$. 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 $$\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{sin(p\pi)}$$ 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.

 

[d_leet]

$$\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{\sin\left(\pi p\right)}$$

If p is an integer, that is obviously false as sin(p*pi)=0.

 

[Count Iblis]

The integral converges for $$0<p<1$$. 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 $$x^{p-1}$$. If we change this to $$x^{p}$$ 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.

 

[mathwonk]

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).

 

[PowerIso]

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.

 

[Gib Z]

The integral $$\int_0^1\frac{ln(1+x)}{x}\cdot dx$$ can be evaluated by considering the series expansion of the numerator, i.e. $$ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\cdot \frac{x^k}{k}$$. Putting this into the integral gives after dividing it by $$x$$, switching the integral and the sum (permitted due to converging series): $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}x^{k-1}dx$$. This is after evaluation: $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}$$. This can be split up into the following two series $$\left(1+\frac{1}{3^2}+\frac{1}{5^2}+...\right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\right)$$.
These are respectively equal to $$\frac{\pi^2}{8}$$ and $$\frac{\pi^2}{24}$$. The result is therefore $$I=\frac{\pi^2}{12}$$. 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 $\frac{\zeta (2)}{2}$ which indeed gives $\pi^2/6$ although the value of zeta 2 is much more well known.

 

[coomast]

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 $$\int_{0}^{1}t^{m-1}(1-t)^{n-1}dt$$. Using the following substitution $$m=1-p$$ and $$n=p$$ gives $$\Gamma(p)\Gamma(1-p)$$, which can be shown to be $$\frac{\pi}{sin(p\pi)}$$. This is done by contour integration. (Details can be found in Spiegel).

The integral of the beta function can be transformed using the substitution $$u=\frac{t}{1-t}$$ 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 $$p=1-y$$, 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. $$sin((1-p)\pi)=sin(p\pi)$$.

The following can also be shown to hold: $$\int_{0}^{\infty}\frac{x^{-p}}{x-1}dx=\pi\cdot cot(p\pi)$$ 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 :-)

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

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

 

[sutupidmath]

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 $$\int_{0}^{1}t^{m-1}(1-t)^{n-1}dt$$. Using the following substitution $$m=1-p$$ and $$n=p$$ gives $$\Gamma(p)\Gamma(1-p)$$, which can be shown to be $$\frac{\pi}{sin(p\pi)}$$. This is done by contour integration. (Details can be found in Spiegel).

The integral of the beta function can be transformed using the substitution $$u=\frac{t}{1-t}$$ 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 $$p=1-y$$, 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. $$sin((1-p)\pi)=sin(p\pi)$$.

The following can also be shown to hold: $$\int_{0}^{\infty}\frac{x^{-p}}{x-1}dx=\pi\cdot cot(p\pi)$$ 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 :-)

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

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

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

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

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

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

 

[sutupidmath]

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 $$\int_{0}^{1}t^{m-1}(1-t)^{n-1}dt$$. Using the following substitution $$m=1-p$$ and $$n=p$$ gives $$\Gamma(p)\Gamma(1-p)$$, which can be shown to be $$\frac{\pi}{sin(p\pi)}$$. This is done by contour integration. (Details can be found in Spiegel).

The integral of the beta function can be transformed using the substitution $$u=\frac{t}{1-t}$$ 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 $$p=1-y$$, 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. $$sin((1-p)\pi)=sin(p\pi)$$.

The following can also be shown to hold: $$\int_{0}^{\infty}\frac{x^{-p}}{x-1}dx=\pi\cdot cot(p\pi)$$ 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 :-)

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

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

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

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

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

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. $$\int[x]dx$$, where x>=0

 

[sutupidmath]

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:

$$\int_0^1\{f'(1-x^{n})x^{2n-1}}\ dx$$ =1/n $$\int_0^1\{f(x)}\ dx$$

 

[Gib Z]

Define [x] please =]? I've seen that to mean either the ceiling or floor functions.

 

[sutupidmath]

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.

 

[coomast]

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 $$1-x^n=y$$ in the left hand side integral.
This gives now $$\frac{1}{n}\int_{0}^{1}f'(y)\cdot (1-y)\cdot dy$$. This can be rewritten using partial integration to give:
$$\frac{1}{n}\int_{0}^{1}f'(y)\cdot dy-\frac{1}{n}\int_{0}^{1}f'(y)\cdot y \cdot dy=\frac{1}{n}(f(1)-f(0))-\frac{1}{n}\left(\left[yf(y)\right]_{0}^{1}-\int_{0}^{1}f(y)\cdot dy\right)$$

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

$$\frac{1}{n}f(1)-\frac{1}{n}f(1)+\frac{1}{n}\int_{0}^{1}f(y)\cdot dy=\frac{1}{n}\int_{0}^{1}f(y)\cdot dy$$

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.

 

[tronter]

$$\int_{0}^{\pi/2} ab \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt$$

 

[sutupidmath]

1. $$\int[x]dx$$, where x>=0

If you guys want i could throw some hints?

 

[sutupidmath]

$$\int_{0}^{\pi/2} ab \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt$$

i thinki got it. I am going to post it very briefly because my friend is calling me up

 

[sutupidmath]

$$\int_{0}^{\pi/2} \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt$$=ab$$\int_{0}^{\pi/2} \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt$$=ab/2 $$\int_{0}^{\pi/2} 2( \sin t \cos t) \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt$$=ab/2 $$\int_{0}^{\pi/2} ( \sin 2t ) \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt$$=ab/2 $$\int_{0}^{\pi/2} ( \sin 2t ) \sqrt{a^{2} \((1-cos 2t)/2 + b^{2} \((1+cos2t)/2} \ dt$$=
then after some other transformaions i come to this point
ab/sqrt2 $$\int_{0}^{\pi/2} ( \(sin 2t)/2 ) \sqrt{cos2t(b^{2}-a^{2})+a^{2}+b^{2}} \ dt$$=
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 $$\int_{2b^{2}}^{\2a^{2}} ({ \sqrt u} \ du$$=

SO now it gets pretty easy to integrate.

 

[sutupidmath]

the upper bound here should be 2a^2

 

[Gib Z]

I'm not too sure what exactly you did, but there is an easier way :(

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

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

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

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

Note: It seems we get different answers :(

 

[Gib Z]

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 :(

 

[sutupidmath]

I'm not too sure what exactly you did, but there is an easier way :(

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

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

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

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

Note: It seems we get different answers :(

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

$$\frac{4(ab)(a^2+ab+b^2)}{3(a+b)}$$
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.

 

[sutupidmath]

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

 

[rocomath]

$$I=\int \frac{\sin \theta - \cos \theta}{(\sin \theta + \cos \theta)\sqrt{\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta}} d \theta$$
$$= \int \frac{\sin^2 \theta - \cos^2 \theta}{(1 + 2\sin \theta \cos \theta)\sqrt{\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta}} d \theta$$

Now let $$u=\sin \theta \cos \theta$$
$$\frac{du}{d\theta}=\cos^2\theta-\sin^2\theta$$

This gives

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

Now let $$l=\sqrt{u+u^2}$$
$$\frac{dl}{du}=\frac{1+2u}{2\sqrt{u+u^2}}$$

This gives
$$I=\int \frac{-2}{4l^2+1}dl$$

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

 

[mathwonk]

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.

 

[Gib Z]

Was that aimed at me or rocophysics? I know I stink :( Whats a wine?

 

[rocomath]

Was that aimed at me or rocophysics? I know I stink :( Whats a wine?

It was aimed at you, I've already spoiled :p

 

[Nebuchadnezza]

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

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

Find the area between the function $f(x)$ and the x-axis when $f(x)=\sqrt{a-\sqrt{x}}$

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

Medium

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

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

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

"Fun"

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

Have fun and gl.