1. ## Some challenge integrals to think about.

$1.) \quad \int_0^1\frac{x^4+1}{x^6+1}dx$

$2.) \quad \int_2^3\frac{x}{x^8+1}dx$

$3.) \quad \int_1^{\sqrt{3}}\frac{x^2+1}{x^4+x^2+1}dx$

$4.) \quad \int_0^3\arcsin{\sqrt{\frac{x}{x+1}}}dx$

$5.) \quad \int_{-1}^1\frac{dx}{(e^x+1)(x^2+1)}$

$6.) \quad \int_{-\pi/4}^{\pi/4}\frac{x^7-3x^5+7x^3-x+1}{\cos^2{x}}dx$

$7.) \quad \int_0^1\left[\ln\left(x+\sqrt{x^2+1}\right)\right]^2dx$

2. ## Re: Some challenge integrals to think about.

Do all of these have exact solutions? That is to say exact without needing 12 terms involving logarithms and arctangents. I had to use the full Mathematica package to do the first two...Wolfram|Alpha just wan't up to it.

-Dan

3. ## Re: Some challenge integrals to think about.

Originally Posted by topsquark
Do all of these have exact solutions? That is to say exact without needing 12 terms involving logarithms and arctangents. I had to use the full Mathematica package to do the first two...Wolfram|Alpha just wan't up to it.

-Dan
Yes, all of them have exact solution and can be integrated using man power. No need to use Mathematica or Wolfram. Our "super-computer" is much more capable and sophisticated than those softwares. In addition, no need to utilize advance concepts such as complex integration etc.... My answer to #1 is $\frac{\pi}{3}$.

4. ## Re: Some challenge integrals to think about.

$\text{Problem 2 should read as:}\quad \int_2^3\frac{x}{x^8-1}dx$

5. ## Re: Some challenge integrals to think about.

The first 3 can all be done with brute force, simply by using partial fractions, breaking the function into terms of the form $\frac{A_n}{x-r_n}$, then integrating term by term.

I used a quicker method for #1 though, and I got $\frac{\pi}{12}\left (\cot\left (\frac{5\pi}{12}\right ) + \tan\left (\frac{5\pi}{12}\right )\right )$. Then I looked up that the answer was Pi/3, and interestingly my answer does break down to pi/3 with the use of some identities.

6. ## Re: Some challenge integrals to think about.

Originally Posted by thevinh
...
Nice questions!

$5.) \quad \int_{-1}^1\frac{dx}{(e^x+1)(x^2+1)}$
Spoiler:

Let:

\displaystyle \begin{aligned} I & = \int_{-1}^{1} \frac{dx}{(e^x + 1)(x^2 + 1)} \overset{x\mapsto -x}= \int_{-1}^{1} \frac{e^x \ dx}{(e^x + 1)(x^2 + 1)} \\ \\ 2I & = \int_{-1}^{1} \frac{dx}{x^2 + 1} \implies I = \frac{\pi}{4} \end{aligned}.

7. ## Re: Some challenge integrals to think about.

$4.) \quad \int_0^3\arcsin{\sqrt{\frac{x}{x+1}}}dx$

Spoiler:
The easiest way to think about this one is to visualize the graph of the function. Then, you can find the area using the inverse of this integrand, which, in fact, is simply tan(x)^2. Once you draw the graph and realize what needs to be done, the integral becomes

$\int_0^\frac{\pi}{3} 3-\tan(x)^2 dx$
$4x - \tan(x)$ evaluated at $x = \frac{\pi}{3}$
$= \frac{4\pi}{3} - \sqrt{3}$

8. ## Re: Some challenge integrals to think about.

Originally Posted by thevinh
$7.) \quad \int_0^1\left[\ln\left(x+\sqrt{x^2+1}\right)\right]^2dx$
Spoiler:

\displaystyle \begin{aligned} \int_0^1 \left[ \ln \left( x + \sqrt{x^2 + 1}\right)\right]^2 \ dx & = \int_0^1 \text{arsinh}^2 \ x \ dx \\ & = \int_0^{\text{arsinh} \1} x^2 \cosh x \ dx \quad \quad \quad\quad\quad (\text{arsinh} \ x\mapsto x) \\ & = \text{arsinh}^2 \ 1 - 2\sqrt{2} \ \text{arsinh} \ 1 + 2 \quad \quad (\text{I.B.P.} \times 2})\end{aligned}

Here are some more:

$8)$ Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^{12} x}{\sin^{12} x + \cos^{12} x} \ dx$

$9)$ Evaluate $\lim_{n \to \infty} \int_1^{n} \frac{n}{1 + x^n} \ dx$.

$10)$ Evaluate $\int_0^{\frac{\pi}{2}} \left( \frac{x}{\sin x}\right)^2 \ dx$

9. ## Re: Some challenge integrals to think about.

I was just finishing that one but you beat me to it

10. ## Re: Some challenge integrals to think about.

Originally Posted by SworD
I was just finishing that one but you beat me to it
Hah, sorry. Ninja'd.

Originally Posted by thevinh
$6.) \quad \int_{-\pi/4}^{\pi/4}\frac{x^7-3x^5+7x^3-x+1}{\cos^2{x}}dx$
Spoiler:

Let:

\displaystyle \begin{aligned} I &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x^7 - 3x^5 + 7x^3 - x + 1}{\cos^2 x} \ dx \overset{x\mapsto -x}= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-(x^7 -3x^5 + 7x^3 - x) + 1}{\cos^2 x} \ dx \\ \\ 2I & = 2 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \ dx \implies I = 2 \end{aligned}

$11)$ Evaluate $\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \ dx$.

11. ## Re: Some challenge integrals to think about.

Originally Posted by FelixFelicis28
Nice questions!
Spoiler:

Let:

\displaystyle \begin{aligned} I & = \int_{-1}^{1} \frac{dx}{(e^x + 1)(x^2 + 1)} \overset{x\mapsto -x}= \int_{-1}^{1} \frac{e^x \ dx}{(e^x + 1)(x^2 + 1)} \\ \\ 2I & = \int_{-1}^{1} \frac{dx}{x^2 + 1} \implies I = \frac{\pi}{4} \end{aligned}.
Just saw this now, thats so witty O.O

12. ## Re: Some challenge integrals to think about.

Originally Posted by SworD
Just saw this now, thats so witty O.O
Thank you! Integration is actually one of my most favourite topics so needless to say, I've had quite a bit of practice haha.

I'm not sure how you or topsquark did number 1 but here's another method:

$1) \ \int_0^1 \frac{x^4 + 1}{x^6 + 1} \ dx$
Spoiler:
\displaystyle \begin{aligned} \int_0^1 \frac{x^4 + 1}{x^6 + 1} \ dx & = \int_0^1 \frac{x^4 + 1}{(x^2 + 1)(x^4 - x^2 + 1)} \ dx \\ & = \int_0^1 \frac{dx}{x^2 + 1} + \int_0^1 \frac{x^2}{x^6 + 1} \ dx \\ & = \frac{\pi}{4} + \int_0^1 \sum_{r \geq 0} (-1)^r x^{6r + 2} \ dx \\ & = \frac{\pi}{4} + \sum_{r \geq 0} \int_0^1 (-1)^r x^{6r +2} \ dx \\ & = \frac{\pi}{4} + \frac{1}{3} \sum_{r \geq 0} \frac{(-1)^r x^{3(2r + 1)}}{2r + 1} \bigg|_0^1 \\ & = \frac{\pi}{4} + \frac{1}{3} \arctan x^3 \bigg|_0^1 \\ & = \frac{\pi}{4} + \frac{\pi}{12} \\ & = \frac{\pi}{3} \end{aligned}

13. ## Re: Some challenge integrals to think about.

Originally Posted by FelixFelicis28
Thank you! Integration is actually one of my most favourite topics so needless to say, I've had quite a bit of practice haha.

I'm not sure how you or topsquark did number 1 but here's another method:

Spoiler:
\displaystyle \begin{aligned} \int_0^1 \frac{x^4 + 1}{x^6 + 1} \ dx & = \int_0^1 \frac{x^4 + 1}{(x^2 + 1)(x^4 - x^2 + 1)} \ dx \\ & = \int_0^1 \frac{dx}{x^2 + 1} + \int_0^1 \frac{x^2}{x^6 + 1} \ dx \\ & = \frac{\pi}{4} + \int_0^1 \sum_{r \geq 0} (-1)^r x^{6r + 2} \ dx \\ & = \frac{\pi}{4} + \sum_{r \geq 0} \int_0^1 (-1)^r x^{6r +2} \ dx \\ & = \frac{\pi}{4} + \frac{1}{3} \sum_{r \geq 0} \frac{(-1)^r x^{3(2r + 1)}}{2r + 1} \bigg|_0^1 \\ & = \frac{\pi}{4} + \frac{1}{3} \arctan x^3 \bigg|_0^1 \\ & = \frac{\pi}{4} + \frac{\pi}{12} \\ & = \frac{\pi}{3} \end{aligned}
Here is how I do #1. I tried to keep things as simple as possible.

\displaystyle \begin{aligned} \int_0^1 \frac{x^4 + 1}{x^6 + 1}dx & = \int_0^1 \frac{(x^4-x^2+1)+x^2}{(x^2 + 1)(x^4 - x^2 + 1)}dx\\ &= \int_0^1 \frac{dx}{x^2 + 1} + \int_0^1 \frac{x^2}{x^6 + 1}dx\\ &=\frac{\pi}{4}+ \int_0^1 \frac{x^2}{x^6 + 1}dx \end{aligned}

Use simple substitution for the second integral.

$\text{Let}~u=x^3 ~ \Rightarrow~ du=3x^2dx$

\begin{aligned} I_2 &=\int_0^1 \frac{x^2}{x^6+1}dx\\ &=\frac{1}{3}\int_0^1 \frac{du}{u^2+1}\\ &=\frac{1}{3} \arctan{(x^3)}\Right |_0^1 \\ &= \frac{\pi}{12} \end{aligned}

Thus,

$\frac{\pi}{4} + \frac{\pi}{12} = \frac{\pi}{3}$

14. ## Re: Some challenge integrals to think about.

Here's how I did it:

$1.) \quad \int_0^1\frac{x^4+1}{x^6+1}dx$

$= \sum_{n=0}^{\infty} \int_{0}^{1} (x^4+1)((-1)^n x^{6n})$

$= \sum_{n=0}^{\infty} \frac{1}{12n+1} - \frac{1}{12n+7} + \sum_{n=0}^{\infty} \frac{1}{12n+5} - \frac{1}{12n+11}$

$= \sum_{n=-\infty}^{\infty} \frac{1}{12n+5} - \frac{1}{12n+11}$ (notice the change in starting index)

Let $f(z) = \left (\frac{1}{12z+5} - \frac{1}{12z+11} \right ) \pi \cot(\pi z)$

The contour integral of this function around a circle of increasing radius approaches 0, so the sum of the residues must be 0. The residues at any integers are exactly the terms in the series, so the sum of the series is the sum of the residues at z=-5/12 and z=-11/12, times negative 1. This is a common approach in complex analysis for summing series.

$= -\left (\frac{1}{12} \pi \cot(\frac{-5}{12} \pi) - \frac{1}{12} \pi \cot(\frac{-11}{12} \pi) \right ) =$
$\frac{\pi}{12}\left (\cot\left (\frac{5\pi}{12}\right ) + \tan\left (\frac{5\pi}{12}\right )\right )$

Now I must admit, compared to the solutions above this method is rather unusual. But it is interesting because it arrives at a result which is impressed in a different form, yet equal to $\frac{\pi}{3}$.