Challenging Integrals in Calculus 1-2: Expand Your Problem-Solving Skills!

[pakmingki]

can someone give me some really hard intergrals to solve?

make sure they are in the range of calculus 1-2 (anything before multivariable)

My teacher assigned some few hard integrals, and they are fun. I want to try moer.
thanks.

 

[yip]

Try \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}}
(forgot to put the integral sign in, it is now fixed)

 

[itsjustme]

sin(2x)cos(2x)dx

 

[ObsessiveMathsFreak]

\int e^{-x^2} dx

 

[VietDao29]

\int e^{-x^2} dx

I doubt that it belongs to either Calculus 1 or Calculus 2 problems.

... make sure they are in the range of calculus 1-2 (anything before multivariable)...

 

[Gib Z]

\int^{1}_0 \frac{\log_e (1+x)}{x} dx. Quite an interesting one that someone gave to me. Nice Solution :)

 

[Kurdt]

Find \frac{f'(x)}{f(x)} where f(x) = sec(x)+tan(x)and hence find \int sec(x) dx.

One of my faves

 

[pakmingki]

wow, these loko pretty fun. THey look way different from the ones I've ever seen.

Ill give them a whirl sometime soon.

 

[bob1182006]

This is a pretty hard one but I haven't finished Calc 2 so I don't know any harder than this.

My favorite Integral so far is this:

\int \frac{dx}{(x^2+9)^3}

It's general form is of
\int \frac{dx}{(x^2+a^2)^n}

It has a really interesting answer

 

[zoki85]

Hard ,but famous and bautiful :

\int_{0}^{\infty}sin(x^2)dx

 

[janhaa]

Try this one...

\int \sqrt{\tan(x)}{\rm dx}

 

[bit188]

Try this one...

\int \sqrt{\tan(x)}{\rm dx}

That's a good one

 

[Invictious]

Hard ,but famous and bautiful :

\int_{0}^{\infty}sin(x^2)dx

Took me 5 minutes only
That question though, however, was just..simply amazing.
I suggest everyone try that question

 

[Gib Z]

I think the original poster has quite enough thanks...he hasn't actually done any of them yet.

 

[JohnDuck]

Hard ,but famous and bautiful :

\int_{0}^{\infty}sin(x^2)dx

I'm stumped but intrigued.

 

[Equilibrium]

\int \frac{1}{x^5+1}dx

 

[zoki85]

Took me 5 minutes only
That question though, however, was just..simply amazing.
I suggest everyone try that question

We are not all as clever as you Invictious

 

[DyslexicHobo]

\int \frac{1}{x^5+1}dx

I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr

I used u-substitution (well, r-substitution), where r = x^5 + 1. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

And please tell us how to do \int_{0}^{\infty}sin(x^2)dx

 

[prasannapakkiam]

\int e^{-x^2} dx

how can this even be integrated?

 

[JohnDuck]

how can this even be integrated?

It can be proved that there's no elementary antiderivative, but you can use a trick from multivariable calculus involving a change to polar coordinates and the squeeze theorem to evaluate it. It's called a Gaussian integral.

Edit: Correction--the trick works for \int_{- \infty}^{\infty} e^{-x^{2}}dx

 

[janhaa]

Integration

how can this even be integrated?

Observe that \,e^{-x^2}\,is an even function, and we can integrate in two dimensions ):

I^2=(\int_{\mathbb{R}} e^{-x^2}){\rm dx})^2=(\int_{\mathbb{R}}e^{-x^2}{\rm dx})(\int_{\mathbb{R}}e^{-y^2}{\rm dy})

I^2=\int_{\mathbb{R}} \int_{\mathbb{R}} e^{-(x^2+y^2)}{\rm dx}{\rm dy}

Then change to polar coordinates:

I^2=\int_0^{2\pi}\int_0^{\infty} e^{-r^2} r {\rm dr} {\rm d\theta}=2\pi \int_0^{\infty} e^{-r^2} r {\rm dr}

then substitution:

\, u = r^2 \,

\frac{\rm du}{2r}={\rm dr}

that is:

I^2=\pi \int_0^{\infty} e^{-u} {\rm du}= \pi

finally:

I=\int_{\mathbb{R}} e^{-x^2} {\rm dx}=\int_{- \infty}^{\infty} e^{-x^2} {\rm dx}=\sqrt{\pi}

 

[Office_Shredder]

Way to drop the ball on the limits at the end...

 

[yip]

I'm stumped but intrigued.

I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr

I used u-substitution (well, r-substitution), where r = x^5 + 1. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

And please tell us how to do \int_{0}^{\infty}sin(x^2)dx

For that integral, i think u have to use the partial fraction theorem that involves derivatives, as there is trigonometric terms in the primitive of that function. Here is how I did the second integral, since u asked, the trick is to apply euler's theorem:

e^{-ix^{2}}=cosx^{2}-isinx^{2}
\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx
It is known that \int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}
\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}
\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}
\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)
\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}

 

[JohnDuck]

For that integral, i think u have to use the partial fraction theorem that involves derivatives, as there is trigonometric terms in the primitive of that function. Here is how I did the second integral, since u asked, the trick is to apply euler's theorem:

e^{-ix^{2}}=cosx^{2}-isinx^{2}
\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx
It is known that \int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}
\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}
\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}
\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)

I follow you up to here.

\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}

Wha?

 

[yip]

Its simply taking the imaginary part of the integral, as the imaginary part of e^-ix^2 is -sinx^2

 

[JohnDuck]

Oh. That makes sense.

 

[DyslexicHobo]

I followed you up to about the part where... uhh nevermind. Didn't catch any of that. :/

Way above my head. Thanks for the explanation, though. I don't even understand how we can even begin to integrate a transcendental function using limits of infinity. They don't have a value at infinity, so how can they be evaluated?

 

[JohnDuck]

Improper integrals such as:

\int_{a}^{\infty} f(x)dx

are defined as such:

\lim_{b\rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{b\rightarrow \infty} F(b) - F(a)

where F(x) is the antiderivative of f(x). If you don't understand how to take limits of functions at infinity, you should probably read up on limits again. Most calculus texts have a brief section on limits, and any introductory real analysis text certainly covers the topic thoroughly.

 

[FlashStorm]

I am scared and frightened.

 

[Kurdt]

I am scared and frightened.

 

[DyslexicHobo]

Improper integrals such as:

\int_{a}^{\infty} f(x)dx

are defined as such:

\lim_{b\rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{b\rightarrow \infty} F(b) - F(a)

where F(x) is the antiderivative of f(x). If you don't understand how to take limits of functions at infinity, you should probably read up on limits again. Most calculus texts have a brief section on limits, and any introductory real analysis text certainly covers the topic thoroughly.

Sorry, I must have mis-worded myself. I meant that I don't understand how trigonometric functions can be evaluated as their inside approaches infinity. The function oscillates between 1 and -1, and never converges. I'm assuming that the Fundamental Theorem cannot be used here befause \lim_{x\rightarrow\infty} 2xcos(x^2) (2xcos(x^2) is the anti-derivative of the starting function) cannot be evaluated, so it seems.

 

[JohnDuck]

Sorry, I must have mis-worded myself. I meant that I don't understand how trigonometric functions can be evaluated as their inside approaches infinity. The function oscillates between 1 and -1, and never converges. I'm assuming that the Fundamental Theorem cannot be used here befause \lim_{x\rightarrow\infty} 2xcos(x^2) (2xcos(x^2) is the anti-derivative of the starting function) cannot be evaluated, so it seems.

That's actually not the antiderivative of \sin{x^{2}}. You're right in that, for example, \lim_{x\rightarrow \infty} \sin{x} doesn't converge in the real numbers. This can be proven pretty easily. However, depending on the function inside of sine, it may converge. Consider \lim_{x\rightarrow \infty} \sin{\frac{1}{x}}--it converges to 0.

Edit: My gut says that \sin{x^{2}} has no elementary antiderivative, but I'm not sure how one might prove or disprove this.

 

[Gib Z]

Well by integrating the Taylor series of sin(x^2) term by term we get:

\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{4n-1}}{(4n-1)(2n+1)!}

which one could recognize as \sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)

where S(u)=\int^u_0 \sin \left(\frac{1}{2} \pi x^2\right) dx.

As S(u) is not an elementary function, we have proved \sin x^2 has no elementary derivative.

 

[yip]

Well by integrating the Taylor series of sin(x^2) term by term we get:

\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{4n-1}}{(4n-1)(2n+1)!}

which one could recognize as \sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)

where S(u)=\int^u_0 \sin \left(\frac{1}{2} \pi x^2\right) dx.

As S(u) is not an elementary function, we have proved \sin x^2 has no elementary anti-derivative.

Isn't that a sort of circular argument? If u assume S(u)=integral of sin(cx^2), where c=pi/2 does not have a closed form answer, aren't u also implicitly assuming that the integral of sin(x^2), the case where c=1, also does not have a closed form answer? I was under the impression that changing constants doesn't really affect integrability, only changing the variables that u are integrating with respect to would. I think that proving that something is not integratable would involve much more complicated arguments. This is just my opinion though. My attempt at saying something is not integrable would rely on the well known fact that e^-ix^2 has no closed form integral, and since the sum of the integral of isin(x^2) and cos(x^2) equals the integral of e^-ix^2, if a closed form integral did exist for sin(x^2), then one must also exist for cos(x^2) (since the cosine function is just a shifted sine function), which means a closed form integral exists for e^-ix^2, which is a contradiction.

 

[Gib Z]

Thats what a lot of non-elementary functions are :) Functions invented with the pure purpose of being the antiderivative of something that otherwise wouldn't have one. eg The Error function, of SI(x)

 

[rocomath]

Try \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}}
(forgot to put the integral sign in, it is now fixed)

is this a Calc 1 or 2 problem?

 

[HallsofIvy]

I would consider it Calc II.

 

[rocomath]

I would consider it Calc II.

i thought so, because i didn't have a prob like that in Calc 1 ... I'm dying to know though, lol

 

[Kummer]

I much more interesting integral is,
\int_0^{\infty} \cos x^k dx = \cos \frac{\pi}{2k} \cdot \Gamma [(k+1)/k]
And,
\int_0^{\infty} \sin x^k dx = \sin \frac{\pi}{2k} \cdot \Gamma [(k+1)/k]

-Wolfgang

 

[Kurdt]

I think the original poster of this thread is only doing them themselves in secret or got scared and ran away.

 

[ansrivas]

\int^{1}_0 \frac{\log_e (1+x)}{x} dx. Quite an interesting one that someone gave to me. Nice Solution :)

I=\int^1_0 \frac{\log(1+x)}{x}\,dx = \int^1_0 \frac{1}{x}\sum_{i=1}^{\infty}\frac{x^i(-1)^{i+1}}{i}

=\int_1^0 \left( 1 - \frac{x}{2} + \frac{x^2}{3} - \ldots \right)\,dx = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \ldots

Using

\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}

also gives:

\sum_{n=1}^{\infty} \frac{1}{(2n)^2}=\frac{\pi^2}{24}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}

finally the desired sum above converges absolutely so we can re-arrange and separate out the +ve and -ve terms giving:

I=\pi^2/8 - \pi^2/24 = \pi^2/12

 

[Klaus_Hoffmann]

\sqrt tan(x) can't be reduced to Rational functions or known functions

 

[ansrivas]

I much more interesting integral is,
\int_0^{\infty} \cos x^k dx = \cos \frac{\pi}{2k} \cdot \Gamma [(k+1)/k]
And,
\int_0^{\infty} \sin x^k dx = \sin \frac{\pi}{2k} \cdot \Gamma [(k+1)/k]

-Wolfgang

I=\int_0^\infty \exp\left( -i x^k \right)

Substituting:

u=ix^k
x\frac{du}{dx}=kix^k=ku

I=\left( \int_0^\infty \exp(-u)u^{1/k-1} \right) \frac{(-i)^{1/k}}{k}

=\frac{1}{k}\Gamma[\frac{1}{k}]\exp\left( \frac{-i\pi}{2k}\right) = \Gamma[(k+1)/k]\exp\left( \frac{-i\pi}{2k}\right)

Comparing Real and Imag parts gives the desired relations.

 

[Gib Z]

ansrivas, nice solutions :) Both are quite correct, of course.

And Klaus, I'm sure you will find that \int \sqrt{ \tan x} is quite possible to do, and is expressible with trig, inverse trig and log functions. The solution can be seen on the integrator, http://integrals.wolfram.com/index.jsp.

If you wish to know how to derive the answer, its quite easy stuff, just takes a lot of time.

After letting tan x = u², u convert it to 2\int \frac{u^2}{1+u^4} du. Complete the square at the bottom to get (u²+1)²-2u² = (u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1). Your integral becomes \int \frac{u^2}{(u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1)} du.

By partial fractions, \frac{u^2}{(u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1)} = \frac{Au+B}{u^2+u\sqrt{2}+1} + \frac{Cu+D}{u^2-u\sqrt{2}+1}. Compare coefficients on both sides to get A=-\frac{1}{2\sqrt{2}}, C=\frac{1}{2\sqrt{2}} and B=D=0.

After that, manipulate the integral by factorization, completing the square and expressing the result in terms of ln and \tan^{-1}, finally u get the answer \frac{\sqrt{2}}{8} ln \frac{u^2-u\sqrt{2}+1}{u^2+u\sqrt{2}+1} + \frac{\sqrt{2}}{4} \tan^{-1} (u\sqrt{2}-1) - \tan^{-1} (u\sqrt{2}+1) + C. The last step is to convert u back to original variable x using the relation u² = tan x and u get your answer in terms of x.

 

[lurflurf]

Gib z cleared this up,but I'll add

\sqrt tan(x) can't be reduced to Rational functions or known functions

sqrt tan(x) is a know function aqt least I know of it.
If you mean the antiderivative or primative there of I will let you know that indefinite integrals of Sqrt tan(x), 1/sqrt tan(x), and curt tan(x) are well known. They are at the level of a first course in calculus except for such courses avoiding messy algebra.

We did it here
https://www.physicsforums.com/showthread.php?p=771121#post771121
where

\int \sqrt{\tan(x)}dx=\int\frac{\sec^2(x)}{1+\tan^2(x)}\sqrt{\tan(x)}dx=\int\frac{2\tan(x)}{1+\tan^2(x)}d\sqrt{\tan(x)}
so it hinges on the always fun
\int\frac{x^2}{1+x^4}dx

 

[Gib Z]

Yes that thread is quite good, the one lurflurf posted. Post 12 has an easier method to do the integral than my post's.

 

[ansrivas]

Another interesting integral I remember from one of Feynman popular book is

\int_0^\pi \log\left( 1-2\alpha \cos x + \alpha^2 \right) dx

However, I can't recall it at all now. I remember it was the same trick as that used to find the definite integral of the sinc function (differentiate a definite integral with some parameter and form a simpler differential equation).

Any ideas?

 

[yip]

Let
I=\int_{0}^{\pi}log(1-2\alpha cosx+\alpha^{2})dx
\frac{dI}{d\alpha}=\int_{0}^{\pi}\frac{-2cosx+2\alpha}{1-2\alpha cosx+\alpha^{2}}
=\frac{1}{\alpha}\int_{0}^{\pi}(1-\frac{1-\alpha^{2}}{1-2\alpha cosx+\alpha^{2}})
=\frac{\pi}{\alpha}-\frac{1}{\alpha}\int_{0}^{\pi}\frac{1-\alpha^{2}}{1+\alpha^{2}-2\alpha\frac{1-tan^{2}\frac{x}{2}}{1+tan^{2}\frac{x}{2}}}
=\frac{\pi}{\alpha}-\frac{2}{\alpha}\int_{0}^{\pi}\frac{1}{2}\frac{\frac{1+\alpha}{1-\alpha}sec^{2}\frac{x}{2}}{1+\frac{(1+\alpha)^{2}}{(1-\alpha)^{2}}tan^{2}\frac{x}{2}}
=\frac{\pi}{\alpha}-\frac{2}{\alpha}[tan^{-1}[(\frac{1+\alpha}{1-\alpha})tan\frac{x}{2}]]_{0}^{\pi}
=\frac{\pi}{2}when -1<\alpha<1, -\frac{\pi}{2}when \alpha<-1, \alpha>1
When -1<\alpha<1, \frac{dI}{d\alpha}=0 so I=k, where k is a constant
Letting \alpha=0, I=0, so I=0 (-1<\alpha<1)
When \alpha<-1, \alpha>1, \frac{dI}{d\alpha}=\frac{2\pi}{\alpha}, I=2\pi log(\alpha)+C where C is a constant. To determine the constant, a value greater than 1 or less than 1 must be substituted, but this is rather messy, so let \alpha=\frac{1}{\beta} where -1<\beta<1
I=\int_{0}^{\pi}log(1-\frac{2}{\beta} cosx+\frac{1}{\beta^{2}})<br /> =\int_{0}^{\pi}log(1-2\beta cosx+\beta^{2})+log(\frac{1}{\beta^{2}})dx
Since we already evaluated the integral when the parameter was between -1 and 1, and found the constant to be 0, likewise, the constant must also be 0 in the above case

When \alpha=1, I=\int_{0}^{\pi}log(2-2cosx)dx=\int_{0}^{\pi}log(4sin^{2}\frac{x}{2})dx
=\pi log4+4\int_{0}^{\frac{\pi}{2}}sinydy,wherey=\frac{x}{2}
\int_{0}^{\frac{\pi}{2}}sinydy=\int_{0}^{\frac{\pi}{2}}cosydy
2\int_{0}^{\frac{\pi}{2}}sinydy=\int_{0}^{\frac{\pi}{2}}log(\frac{sin2y}{2})=\int_{0}^{\frac{\pi}{2}}log(\frac{sin2y})-log2
Since \int_{0}^{\frac{\pi}{2}}log(sin2y)dy=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}log(sinz)dz
=2\int_{0}^{\frac{\pi}{2}}log(sinz)dz, I=\pi log4-4\frac{\pi}{2}log2=0
It can be seen that the exact same working applies when \alpha=-1so finally
I=2\pi log(\alpha^{2}) when (\alpha&lt;-1, \alpha&gt;1) and 0 when (-1\le\alpha\le1)

 

[ansrivas]

Thanks yip. I was stumped on that one for some time.

 

[Gib Z]

For |u| &lt; 1,

\log (1-2u \cos x + u^2) = -2 \sum_{n=1}^{\infty} \frac{u^n}{n} \cos nx

Therefore:

\int^{\pi}_0 \log (1-2u \cos x + u^2)= 2\pi \log |u| \\ \mbox{if} |u| &gt; 1, \\ \mbox{ 0 otherwise}