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

[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

 

[royjsky]

Hard ,but famous and bautiful :

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

I believe this uses multivariable calculus since I recall you must first prove the definite integral of the e^-x^2 which requires double integrals and a change of variables.

 

[royjsky]

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

I think you need multivariable to do that.

 

[Corbeau]

Here is a hard nut to crack. Been trying to solve it for a while, can't confirm it's actually solvable.

1/ ( x ( ln (x+1) -1 ) )

 

[suprabhe]

x^2arctan(5x)

 

[Demystifier]

https://www.amazon.com/dp/1493912763/?tag=pfamazon01-20

 

[MAGNIBORO]

$$\int_{0}^{\pi}\frac{cos(nx)-cos(na)}{cos(x)-cos(a)} dx$$

and

$$\int_{0}^{\infty }\frac{x}{e^{x}-1} dx$$

 

[Suraj Pal]

Try this.

 

[Ritam Dutta]

Hard ,but famous and bautiful :

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

it should come from common sense i think. seems the analytical method is going to be just WOOW

 

[ComplexVar89]

https://www.amazon.com/dp/1493912763/?tag=pfamazon01-20

I was just thinking this thread is quite reminiscent of that book (without Nahin's witty commentary, of course.)

 

[Ferhat]

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)

This is my answer, tell me if i did something wrong :).

 

[quarantine2020]

This is my answer, tell me if i did something wrong :).

this is a “simpler” result I got from unraveling your solution & re-packaging it. I want to figure out how to get it in this form in a more natural way. As of now I’m stuck. Right now I’m working with a Pythagorean triangle
Adjacent = 1-x^2
Opposite = x*sqrt(2)
Hypotenuse = sqrt(1+x^4)

and the solution is 1/sqrt(2)*ln(sec(angle)+tan(angle)) + C

I see some kind of pattern here but it’s a little opaque. Any way to clear this up & produce a really elegant solution?

 

[Ssnow]

Try:

\int_{0}^{1} e^{-x^{x}} dx

Ssnow

 

[akil]

try the integral of sin(lnx) by using eulers formula

 

[Aritro Biswas]

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)

The answer is
<br /> \frac{1}{2\sqrt{2}}\ln \left| { \frac{\sqrt{2}+{\sqrt{x^2+\frac{1}{x^2}}}} {\sqrt{2}-{\sqrt{x^2+\frac{1}{x^2}}}}} \right|<br />

 

[Ossi]

Hard ,but famous and bautiful :

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

Is It -sqrt(pi/2) took me a bit to calculate. Its doable, If one knows the tricks

 

[mathhabibi]

Try this one:
$\int_0^\infty\frac{\sin^2x}{x^2(x^2+1)}dx$
If you need an explanation, let me know. But I want to give you guys some time to find out how to do it

 

[Vanadium 50]

As my first calculus teacher said, "there is a difference between a hard problem and a long problem."

 

[mathhabibi]

As my first calculus teacher said, "there is a difference between a hard problem and a long problem."

Is this directed toward my integral? If it is, I could take it down from this thread.

 

[mathhabibi]

Is this directed toward my integral? If it is, I could take it down from this thread.

Actually, I can't delete that post.

 

[mathhabibi]

Here's another integral that I find interesting $$\int_0^\infty\frac{\sin x}{\sinh x}dx$$This one has an answer in terms of hyperbolic cotangent.