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

[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 $$-1<\beta<1$$
$$I=\int_{0}^{\pi}log(1-\frac{2}{\beta} cosx+\frac{1}{\beta^{2}}) =\int_{0}^{\pi}log(1-2\beta cosx+\beta^{2})+log(\frac{1}{\beta^{2}})dx$$
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 $$\alpha=1, I=\int_{0}^{\pi}log(2-2cosx)dx=\int_{0}^{\pi}log(4sin^{2}\frac{x}{2})dx$$
$$=\pi log4+4\int_{0}^{\frac{\pi}{2}}sinydy,$$where$$y=\frac{x}{2}$$
$$\int_{0}^{\frac{\pi}{2}}sinydy=\int_{0}^{\frac{\pi}{2}}cosydy$$
$$2\int_{0}^{\frac{\pi}{2}}sinydy=\int_{0}^{\frac{\pi}{2}}log(\frac{sin2y}{2})=\int_{0}^{\frac{\pi}{2}}log(\frac{sin2y})-log2$$
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 $$\alpha=-1$$so finally
$$I=2\pi log(\alpha^{2})$$ when $$(\alpha<-1, \alpha>1)$$ and 0 when $$(-1\le\alpha\le1)$$

 

[ansrivas]

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

 

[Gib Z]

For $|u| < 1$,

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

Therefore:

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

 

[arie0003]

Hii,

I found that this topic about calculus is really interesting..
I suggest that we could continue this topic, of course with some new problems. Besides, anyone who could do the integration should put the solution, except the one who purpose the question itself. So, we can improve our skill together. How ?? I hope that everyone have the same thought. Hehe.
Thanks.

 

[Gib Z]

The idea you have is already on this thread
https://www.physicsforums.com/showthread.php?t=149706&page=16

Please post there from now on, and if a mod sees this please lock this thread (and perhaps give the linked thread a sticky position)

 

[jostpuur]

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 $$\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}$$
$$\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}$$
$$\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}$$
$$\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)$$
$$\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$

This is oversimplifying the proof. The usual trick where gaussian integral is calculated as a square root of the two dimensional gaussian integral, gives the equation
$$\int\limits_{-\infty}^{\infty} e^{-(A+iB)x^2}dx = \sqrt{\frac{\pi}{A+iB}}$$
in which A and B are real, only when $A > 0$. (The square root is the one that is continuous on set $\{z\in\mathbb{C}\;|\; \textrm{Re}(z)>0\}$, and that agrees with the positive square root on real axis.) If $A=0$, then the substitution $r=\infty$ does not converge in the two dimensional integral, and thus the result is not trivially the same.

However, the Leibniz's rule of alternating series can be used to show that integrals
$$\int\limits_{0}^{\infty}\cos(x^2)dx$$
and
$$\int\limits_{0}^{\infty}\sin(x^2)dx$$
converge towards something, and thus the integral
$$\int\limits_{-\infty}^{\infty} e^{ix^2}dx$$
also exists. If we know that the integral is a continuous function of the coefficent in the exponent, then this is all done, but I never bothered actually try to figure out how the continuity could be proven. If it can be done nicely, feel free to mention how.

 

[Kummer]

However, the Leibniz's rule of alternating series can be used to show that integrals
$$\int\limits_{0}^{\infty}\cos(x^2)dx$$
and
$$\int\limits_{0}^{\infty}\sin(x^2)dx$$
converge towards something, and thus the integral
$$\int\limits_{-\infty}^{\infty} e^{ix^2}dx$$
also exists. If we know that the integral is a continuous function of the coefficent in the exponent, then this is all done, but I never bothered actually try to figure out how the continuity could be proven. If it can be done nicely, feel free to mention how.

In fact it is known what they converge http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158083"

 

[jostpuur]

The approach that starts with Gaussian integrals leads to the question about limit of the expression
$$\int\limits_{-\infty}^{\infty} e^{-Ax^2}dx$$
when A approaches some value on the line $\{z\in\mathbb{C}\;|\; \textrm{Re}(z)=0\}$ from right.

I admit, I hadn't seen the gamma function approach earlier, and it seems to avoid this problem.

 

[jostpuur]

Or actually I'm not so sure about this all yet. The derivation that uses gamma functions seems to assume right in the beginning that integral
$$\int_{0}^{\infty} e^{ix^k} dx$$
converges. It would be nice to show it some Leibniz's test, for convergence, before continuing with variable changes, that are used for solution of the value of the integral.

 

[Kummer]

Or actually I'm not so sure about this all yet. The derivation that uses gamma functions seems to assume right in the beginning that integral
$$\int_{0}^{\infty} e^{ix^k} dx$$
converges. It would be nice to show it some Leibniz's test, for convergence, before continuing with variable changes, that are used for solution of the value of the integral.

I believe convergence exists when k>1.

 

[iDeriveintegr]

I'm stumped but intrigued.

Isn't it just 0?

 

[DavidAlan]

$$\Gamma(z)=\underset{0}{\overset{\infty}{\int}}\gamma^{z-1}e^{-\gamma}d\gamma$$

Make like a snob and use the gamma-function.

 

[camilus]

$$\int {sinxcosx \over sin^4x+cos^4x}dx$$

 

[dextercioby]

$$I =\int {sinxcosx \over sin^4x+cos^4x}dx$$

Is that supposed to be difficult, or what ?

$$I= -\frac{1}{2}\int \frac{d(\cos 2x)}{(\cos 2x)^2 + 2\left[1-(\cos 2x)^2\right]} =...$$

ends up in something proportional to argth(fraction involving arccos).

 

[camilus]

double angle identity for sine allowed me to solve in terms of arctangent when u=cos(2x)

 

[dextercioby]

Did you get arctangent circular or arctangent hyperbolic ?

 

[camilus]

circular

 

[dextercioby]

it's wrong, because there's a minus in the denominator. it should be the hyperbolic artangent function.

 

[camilus]

i doubt it. Even wolfram integrator says its correct. Show your work?

 

[vector22]

here is a sexy one

\\int \\sqrt {\\x^2+1} dx whoops latex is rusty ... be patient dang how thu??

 

[jhae2.718]

You can use tex tags to display $\LaTeX$.

For the integral, do a trig sub.

 

[gb7nash]

erf!

http://mathworld.wolfram.com/Erf.html

 

[vector22]

This one can be done without the hyperbolic functions but it is a good page long

$$\int \sqrt {x^2 + 1}\, dx$$