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

[AdityaDev]

I need a numerical answer to question in post #135. Its not easy.
I have really difficult questions. The one at #135 is the easiest of them.
Don't compute the indefinite integral. The tricky part is the limits.

 

[AdityaDev]

how? on the left integral, from the beta function you will get gamma functions and no limit problem there.

on the right side there is no limit problem either because it will become integral (ln (u)) which will integrate to u*ln(u)-u, and replacing u=sinx and it's trivial.

Ok. Substitute limits and post the answer.

 

[NeOH]

Ok. Substitute limits and post the answer.

log(2) which is about 0.693 . have a good day

 

[AdityaDev]

Level 2 then
$\int^{\pi}_0\frac{dx}{1-2acosx+a^2}$

 

[AdityaDev]

Or this one
$\int_0^{\pi/2}sinxlnsinxdx$

Now you can't split the integral.

 

[AdityaDev]

I know the answer to both questions.

 

[NeOH]

yeah I am not going to do your homework :P
but I can encourage you by telling that they are simple, and if you know how to get log2, you will know how to solve the sinx lnsinx dx integral because it is part of the previous problem

 

[AdityaDev]

You solved them using beta or gamma functions which we are not taught in school. In school level, if you did not know such functions, then it will be difficult to solve such problems. And don't say they are easy. These questions are made by a mathematician who has a PhD in maths, which you probably don't have. And the level of questions are not easy.

Also, these are not homework questions. If you want you can try them out :P
Its for those who like solving integration problems.

 

[HomogenousCow]

You solved them using beta or gamma functions which we are not taught in school. In school level, if you did not know such functions, then it will be difficult to solve such problems. And don't say they are easy. These questions are made by a mathematician who has a PhD in maths, which you probably don't have. And the level of questions are not easy.

Also, these are not homework questions. If you want you can try them out :P
Its for those who like solving integration problems.

You don't need special functions to solve the integrals.
Yes the limits are not directly computable, however simply taking their limits works.

And, why are you so hostile?

 

[AdityaDev]

I was trying to make things interesting. I like competion irrespective of success or failure.

 

[Mariuszek]

$$\int{\sec^{3}{\theta}\mbox{d}\theta}$$
If you use $$u=\csc{\theta}$$ substitution you will have to do the partial fraction decomposition
$$u=\sec{\theta}+\tan{\theta}$$ will also work
If you prefer reduction formula use it

 

[PWiz]

$$\int{\sec^{3}{\theta}\mbox{d}\theta}$$
If you use $$u=\csc{\theta}$$ substitution you will have to do the partial fraction decomposition
$$u=\sec{\theta}+\tan{\theta}$$ will also work
If you prefer reduction formula use it

This one's super easy (I encountered it while trying to find the arc length of $y=x^2$).
(1) $\int sec^3 \theta\ d\theta = \int sec^2\theta\ sec \theta\ d\theta$. Integrating by parts, we get
(2) $$sec \theta\ tan\theta - \int sec\theta\ tan^2 \theta\ d\theta=sec\theta\ tan\theta - \int sec^3\theta\ d\theta +\int sec\theta\ d\theta$$
So (3) $$\int sec^3 \theta\ d\theta = \frac1{2}(sec\theta\ tan\theta + \int sec\theta\ d\theta)$$
Let's solve the integral of $sec\theta$ on the right. Multiplying the denominator and numerator by $cos \theta$, we get
(4) $$\int \frac{cos \theta}{cos^2 \theta} d\theta = \int \frac{cos \theta}{1-sin^2 \theta} d\theta$$
Factorizing the denominator, we get (5)$$\int \frac{cos \theta}{(1+sin \theta)(1-sin\theta)} d\theta$$
This can be rationalized to give (6)$$\int \frac1{2} (\frac{cos \theta}{1+sin\theta}+\frac{cos \theta}{1-sin\theta}) d\theta$$
Let $u=sin \theta$. Then $du=cos\theta\ d\theta$. So our integral expression becomes from (4) becomes $$\frac1{2}(ln|1+sin\theta|-ln|1-sin\theta|)$$,which is the same as writing $$ln \sqrt{\frac{1+sin \theta}{1-sin\theta}}$$ If we multiply the numerator and denominator by $1+sin\theta$, we get the lovely expression
$$ln \sqrt{\frac{(1+sin\theta)^2}{(cos \theta)^2}}=ln|\frac{1+sin\theta}{cos \theta}|=ln|sec\theta+tan\theta|$$
Making the final substitution into (3), we get $\int sec^3 \theta\ d\theta = \frac1{2}(sec\theta\ tan\theta + ln|sec\theta +tan\theta|)+c$, where $c$ is an arbitrary constant. IMO, this is too trivial a question to appear in this thread. If you really want to push things up a notch though, try the double integral of $sec^3x$ (It uses elliptical integrals, and I myself don't have the slightest clue how to solve it)

 

[PWiz]

Or this one
$\int_0^{\pi/2}sinxlnsinxdx$

Now you can't split the integral.

This is an improper integral. Calculating the indefinite integral is very easy, and this question has more to do with limits than the actual process of integration. In general, using approximate methods is suitable to calculate the numerical value of such functions (although if you want to get very specific, then the question is not correctly presented, as the upper limit is not part of the integrand's domain [a limit must be shown in formal treatment]).

 

[pwsnafu]

This one's super easy (I encountered it while trying to find the arc length of $y=x^2$).
(1) $\int sec^3 \theta\ d\theta = \int sec^2\theta\ sec \theta\ d\theta$. Integrating by parts, we get

Well, if we are discussing $\int \sec^3 (\theta) d\theta$ then there is also
$\int \frac{1}{\cos^3(\theta)} d\theta = \int\frac{\cos(\theta)}{\cos^4(\theta)} d\theta = \int \frac{\cos(\theta)}{(1-\sin^2(\theta))^2} d\theta = \int \frac{1}{(1-u^2)^2} du$

The last integral is tedious but not hard (I've seen it given as a Calc 2 assignment question).

IMO, this is too trivial a question to appear in this thread.

Unfortunately the OP restricted the thread to Calc 1-2 level. There's only so much variation of the basic techniques you can do.

 

[Emmanuel_Euler]

integral from zero to x sin(x^2) is called (fresnel integral) .
the full solution is here.

 

[PWiz]

A must-do preliminary exercise:
$$\int arcsin(x)\ arccos(x) dx$$

 

[certainly]

A must-do preliminary exercise:
$$\int arcsin(x)\ arccos(x) dx$$

That's really easy if you remember that :-
$arcsin(x)+arccos(x)=\frac{\pi}{2}$
then substitute $x=sin(y)$ to get:-
$\int y(\frac{\pi}{2}-y)cos(y) dy$
Here's a good proof of the Gaussian integral that I quite like because at first sight it seems that something is wrong, but the proof is mathematically sound.
Let $I$ be the Gaussian integral then we have:-
$I^2=\int_0^{\infty}e^{-x^2}\int_0^{\infty}e^{-u^2} du dx=\int_0^{\infty}e^{-x^2}\int_0^{\infty}e^{-x^2y^2} x\ dy dx\\ =\int_0^{\infty}\int_0^{\infty} xe^{-x^2(1+y^2)} dx dy\\ =\frac{1}{2}\int_0^{\infty}\frac{1}{1+y^2} dy\\ =\frac{\pi}{4}.$

 

[Alex299792458]

∫ f(x) dx = F(a) - F(b)

 

[pwsnafu]

∫ f(x) dx = F(a) - F(b)

The left hand side is the indefinite integral of f, but the right hand side is a constant. The only function that integrates to a constant is the zero function.

 

[certainly]

If you really want to push things up a notch though, try the double integral of sec3xsec^3x (It uses elliptical integrals, and I myself don't have the slightest clue how to solve it)

This integral is not expressible in terms of elementary functions.
$\iint sec^3(x) dx\ dx=\int\frac{1}{2}[sec(x) tan(x)+ln(sec(x)+tan(x))]+c\ dx\\ =\frac{1}{2}sec(x)+cx+\frac{1}{2}\int ln(sec(x)) dx +\frac{1}{2}\int ln(1+sin(x))dx$
Now note that:- $1+sin(x)=2cos^2(\frac{\pi}{4}-\frac{x}{2})\\ \therefore ln(1+sin(x)=ln (2) -2 ln(sec(\frac{\pi}{4}-\frac{x}{2}))$
and with a simple substitution this integral should be of the $ln(sec(x))$ type.
However $I=\int ln(sec(x)) dx$ is not elementary, see this.

 

[PWiz]

This integral is not expressible in terms of elementary functions.
$\iint sec^3(x) dx\ dx=\int\frac{1}{2}[sec(x) tan(x)+ln(sec(x)+tan(x))]+c\ dx\\ =\frac{1}{2}sec(x)+cx+\frac{1}{2}\int ln(sec(x)) dx +\frac{1}{2}\int ln(1+sin(x))dx$
Now note that:- $1+sin(x)=2cos^2(\frac{\pi}{4}-\frac{x}{2})\\ \therefore ln(1+sin(x)=ln (2) -2 ln(sec(\frac{\pi}{4}-\frac{x}{2}))$
and with a simple substitution this integral should be of the $ln(sec(x))$ type.
However $I=\int ln(sec(x)) dx$ is not elementary, see this.

I know that. It's why I mentioned in my earlier post that it's expressed in terms of "elliptical integrals" and that I can't solve it. In general, I solve out every integral I can on my own unless it's non-elementary, in which case I don't even make an attempt.

 

[certainly]

I know that. It's why I mentioned in my earlier post that it's expressed in terms of "elliptical integrals" and that I can't solve it. In general, I solve out every integral I can on my own unless it's non-elementary, in which case I don't even make an attempt.

I don't see how $I=\int ln(sec(x)) dx$ can be reduced to an elliptic integral...
[EDIT:- remember that elliptic integrals are of the form $\int\frac{A(x)+B(x)\sqrt{S(x)}}{C(x)+D(x)\sqrt{S(x)}} dx$]

 

[PWiz]

I don't see how $I=\int ln(sec(x)) dx$ can be reduced to an elliptic integral...
[EDIT:- remember that elliptic integrals are of the form $\int\frac{A(x)+B(x)\sqrt{S(x)}}{C(x)+D(x)\sqrt{S(x)}} dx$]

You're right, it uses the polylogarithmic function. But the fact that it's non-elementary is enough to keep me from trying

 

[ZetaOfThree]

Try this one $$\int_{0}^{\pi/2}{ dx \over {1+\tan^{2015}{x}}}$$

 

[Alex299792458]

Why specifically to the power of 2015?

 

[maka89]

Here are some I like. First one is one that I stumbled upon while trying to integrate $\exp(-a_1^2x^2 - a_2^2y^2)$ over a circular area.
The second is one that I found was really cool and educational to figure out by myself

Try this:
$\int_0^{2\pi} \exp(-a\cos^2\theta)dx$.

Or this:
$\int \exp(ax)\sin(bx) dx$.

Hint for no.1:
$\int_0^{2}\frac{\exp(-px)}{\sqrt{x(2-x)}}dx = \pi\exp(-p)I_0(p)$.

 

[tommik]

$$\frac{(1-x)^2 (1-x)^2}{(1-x)^2 + 2x} = (1-x)^2 - \frac{2x(1-x)^2}{(1-x)^2+2x} = (1-x)^2 - 2x\frac{1-x^2}{1+x^2} + \frac{4(1+x^2)-4}{1+x^2}$$

The integration of the 4 terms is elementary.

$$\frac{(1-x)^2 (1-x)^2}{(1-x)^2 + 2x} = (1-x)^2 - \frac{2x(1-x)^2}{(1-x)^2+2x} = (1-x)^2 - 2x\frac{1+x^2}{1+x^2} + \frac{4(1+x^2)-4}{1+x^2}$$

I think this is correct

 

[certainly]

$$\frac{(1-x)^2 (1-x)^2}{(1-x)^2 + 2x} = (1-x)^2 - \frac{2x(1-x)^2}{(1-x)^2+2x} = (1-x)^2 - 2x\frac{1+x^2}{1+x^2} + \frac{4(1+x^2)-4}{1+x^2}$$

I think this is correct

Indeed, you are correct.

 

[tommik]

Hi guys...In this 3D I found a lot of interesting posts with amazing solution integrals, other difficult but very boring and other too easy to be interesting. One of the most amazing I found is the following:

$$\int{{\sqrt{tan x}dx}}$$

Here is an example of easy but nice to do:

$$\int{ dx \over {x\sqrt{x^2-a^2}}}$$

$$(-1/a)\int{ -a\over {x|x|\sqrt{1-|a/x|^2}}}dx$$

$$(-1/a)\int{ 1\over {\sqrt{1-|a/x|^2}}}d(|a/x|)= (-1/a) arcsin|a/x| +C$$

 

[ZetaOfThree]

Why specifically to the power of 2015?

You should try the integral and find out for yourself!

 

[certainly]

Try this:
$\int_0^{2\pi} \exp(-a\cos^2\theta)dx$.

NOTE:-To anyone attempting this one, you will need the hint unless you already know about Bessel functions.

Try this one $$\int_{0}^{\pi/2}{ dx \over {1+\tan^{2015}{x}}}$$

The first integral of this type on this thread I believe... the answer is $\frac{\pi}{4}$

One of the most amazing I found is the following:

$\int\sqrt{tan(x)} dx$​

'tis been already posted on this thread.....

Or this:
$\int \exp(ax)\sin(bx) dx$.

'tis easy...just do integration by parts twice on $e^{ax}$...
Come on lads give us something good......[like that $sin(x^2)$ integral, that was simply amazing......]

 

[tommik]

'tis been already posted on this thread.....

@certainly...that's exactly what I said...It is one of the most amazing I found HERE!

 

[phion]

$\int{{\sqrt{1-e^{2}sin^{2}(\theta)} d\theta}}$

 

[phion]

Hint: it's an elliptic integral of the second kind.

 

[phion]

This one might be a little more approachable...

$\int{ x^2-3x+7\over {(x^2-4x+6)^2}}dx$