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.