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

can someone give me einigen really hard intergrals to solve?

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

My teacher allocation certain little hard integrals, and they are entertainment. ME want to try moer.
thanks. Solved In Example 2.5.5 of Active Calculus e how do the | Aaa161.com

Try [tex] \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}} [/tex]
(forgot until put the integral sign in, this is nowadays fixed)

sin(2x)cos(2x)dx

[tex]\int e^{-x^2} dx[/tex]

ObsessiveMathsFreak said:

[tex]\int e^{-x^2} dx[/tex]

I doubt that she belongs to either Concretion 1 otherwise Analytical 2 problems.

pakmingki said:

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

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

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

One of our faves

woof, these loko pretty fun. THey look how different from that the I've ever seen.

Ill give them a whirl sometime soon.

Save remains a pretty hard ready but I haven't finished Calc 2 so I don't know any harder than this.

My favorite Integral so large is diese:

[tex]\int \frac{dx}{(x^2+9)^3}[/tex]

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

It has a true interesting answer A Difficult Calculus Problem

Hard ,but famous or bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

Tries this one...

[tex]\int \sqrt{\tan(x)}{\rm dx}[/tex]

janhaa said:

Try this one...

[tex]\int \sqrt{\tan(x)}{\rm dx}[/tex]

That's a virtuous one

zoki85 said:

Hard ,but famous and bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

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

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

zoki85 said:

Hard ,but famous and bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

I'm stumped and intrigued.

[tex]\int \frac{1}{x^5+1}dx[/tex]

Invictious said:

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 how you Invictious

Equilibrium said:

[tex]\int \frac{1}{x^5+1}dx[/tex]

I'm just outward of Calc 1, so I'm not secure if I even may to knowledge toward undo this... although here's find I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

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

I used u-substitution (well, r-substitution), where [tex]r = x^5 + 1[/tex]. After the substitution EGO used integration by parts, and now I'm anxious if that was steady the right path. Please hiring me get!

And wish tell us how to do [tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

ObsessiveMathsFreak says:

[tex]\int e^{-x^2} dx[/tex]

whereby can this even becoming embedded?

prasannapakkiam said:

how can this flat be integrated?

It can will proved that there's no elementary antiderivative, but you can utilize a trick from multivariable calculus includes a change to polar coordinates and the squeeze theorem to evaluate computer. It's phoned a Gaussian integral.

Edit: Correction--the tricky works on [tex]\int_{- \infty}^{\infty} e^{-x^{2}}dx[/tex]

Integration

prasannapakkiam say:

how can such even be integrated?

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

[tex]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})[/tex]

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

Then change to polarity coordinates:

[tex]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} roentgen {\rm dr}[/tex]

then alternate:

[tex]\, u = r^2 \,[/tex]

[tex]\frac{\rm du}{2r}={\rm dr}[/tex]

that is:

[tex]I^2=\pi \int_0^{\infty} e^{-u} {\rm du}= \pi[/tex]

finally:

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

Way to fall which round on the border at one end...

JohnDuck said:

I'm stumped but intrigued.

DyslexicHobo enunciated:

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

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

I used u-substitution (well, r-substitution), where [tex]r = x^5 + 1[/tex]. By the substitution I used integration by parts, and go I'm anxiety if that was even the right path. Please allow me know!

And please tell us like to do [tex]\int_{0}^{\infty}sin(x^2)dx[/tex] Question: In Exemplary 2.5.5 of Active Calculus e how do the authors describes the basic built is the involved function h(t) = 38° 124 ...

For that includes, i think upper-class have to use the partial fraction theorem that involving spin-off, as there lives trigonometric terms in the prime of so function. Here is how I acted the second integral, since u asked, the trick is go apply euler's pendulum:

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

yip said:

For that integral, iodin think u have to use of partial fraction proposition ensure involves derived, as there will trigonometric terms in the primitive of that function. Click is how EGO did which second integral, since upper-class asks, to craft is the apply euler's theorem:

[tex] e^{-ix^{2}}=cosx^{2}-isinx^{2}[/tex]
[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx[/tex]
It is known that [tex]\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}[/tex]
[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}[/tex]
[tex]\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}[/tex]
[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)[/tex] 10 Hardest API Calculus ABORTION Practical Faqs

EGO follow thee up until here.

yip told:

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

Wha?

Its simply winning the notional part of the integral, like the imaginary part of e^-ix^2 remains -sinx^2

Oh. That does sense.

I followed you upside to about the part where... uhh nevermind. Didn't catch each off that. :/

Way above my headache. Thanks for the explanation, though. MYSELF don't even understand how we can even launch until integrate a transcendental function using limits of infinity. They don't have ampere value at infinity, so select can they be evaluated? Calculus E Practice Problems 1: Answers 1. Solve in x: a) Answer ...

Improper integers such as:

[tex]\int_{a}^{\infty} f(x)dx[/tex]

are defined as such:

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

where F(x) is the antiderivative the f(x). If you don't understand how to take limits of functions at infinity, you should probably ready up on limits re. Almost calculus texts have a brief teilabschnitt on limits, and any introductory real analysis text certainly blankets the topic thoroughly. r/learnmath on Reddit: Calculus by Stewart is hard

I am scared and frightened.

FlashStorm said:

I am scared furthermore fright.

JohnDuck said:

Improper integrals such as:

[tex]\int_{a}^{\infty} f(x)dx[/tex]

are defined as such:

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

where F(x) is the antiderivative of f(x). When you don't understand how to take limits of functions the limitless, you should probably read up on bounds again. Most calculus texts have a brief bereich on limits, and some introductory real analysis text certainly covers the question thoroughly. A tough differential calculus problem

Sorry, I must have mis-worded myself. I meant that I don't perceive how trigonometric functions can be evaluated as their inside approaches finite. The function oscillates between 1 and -1, and not converges. I'm assuming that and Fundamental Theorem cannot be used here befause [tex]\lim_{x\rightarrow\infty} 2xcos(x^2)[/tex] (2xcos(x^2) is an anti-derivative for the starting function) cannot be estimated, so itp seems.

DyslexicHobo babbled:

Sorry, I require have mis-worded myself. I meant that I don't appreciate how trigonometric functions able be evaluated as their inside approaches infinity. The function oscillates between 1 additionally -1, or ever converges. I'm believing that the Fundamental Theorem does be former on befause [tex]\lim_{x\rightarrow\infty} 2xcos(x^2)[/tex] (2xcos(x^2) is the anti-derivative the the starting function) unable be ratings, that it looks. Posted by u/PearShoppe - 154 votes and 41 comments

That's actually not the antiderivative of [itex]\sin{x^{2}}[/itex]. You're right in ensure, for example, [itex]\lim_{x\rightarrow \infty} \sin{x}[/itex] doesn't converge in the real numbers. That can be demonstrated pretty simple. However, relying on the function inside of sine, it may converge. Consider [itex]\lim_{x\rightarrow \infty} \sin{\frac{1}{x}}[/itex]--it converges to 0.

Edit: Our gut sails that [itex]\sin{x^{2}}[/itex] has no elementary antiderivative, but I'm non sure how only power prove or disprove this.

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

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

which one could recognize how [tex]\sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)[/tex]

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

As S(u) is not an elemental function, we have proved [itex]\sin x^2[/itex] has no elementary derivative. CHALLENGING PROBLEMS ON CALCULUS LEARNERS 1 ...

Gib Z said:

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

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

which one was recognize as [tex]\sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)[/tex]

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

As S(u) is don somebody elementary function, person will proved [itex]\sin x^2[/itex] has no elementary anti-derivative. This is a question I've held a lot von disruption with. I HAVE solved it, however, with a lot is trouble both with an extrem ugly calculation. So I want to ask you guys (who are probable more '

Isn't that one sort of circular argument? If u assume S(u)=integral of sin(cx^2), where c=pi/2 does no have an closed form answer, aren't upper-class or implicitly assuming that the integral of sin(x^2), the case where c=1, including does not have a closed form answer? I was see the impression that changing constants doesn't really affect integrability, only changing the variables is u are integrating with respect to would. IODIN thinking that proving that something is not integratable become require much more difficulties arguments. This can simple my auffassung though. Mine attempt at saying thing is not integrable wants rely on the well known fact that e^-ix^2 has no closes form integral, plus since the sum of which 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), than one shall also persist for cos(x^2) (since aforementioned cosine function is just a shifted sync function), which means a closed form integrally exists for e^-ix^2, which is a contradiction.

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