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

[dumbperson]

This one is fairly straightforward, use the substitution $x=\tan\theta$.

For this one, define:
$$I(a)=\int_0^1 \frac{x^a-1}{\ln(x)}dx$$
Differentiate both the side with respect to a to get:
$$\frac{dI}{da}=\int_0^1 x^a\,dx=\frac{1}{a+1}$$
$$\Rightarrow I(a)=\ln|a+1|+C$$
It can be easily seen that C=0. We need the value of I(a) when a=1, hence,
$$I(1)=\ln(2)$$

We use the series expansion of $\ln(1-x)$ i.e
$$\ln(1-x)=-\sum_{k=1}^{\infty} \frac{x^k}{k}$$
Hence, our integral is:
$$-\int_0^1 \frac{1}{x}\sum_{k=1}^{\infty} \frac{x^k}{k}\,dx$$
$$=-\sum_{k=1}^{\infty} \int_0^1 \frac{x^{k-1}}{k}\,dx$$
$$=-\sum_{k=1}^{\infty} \frac{1}{k^2}=-\zeta(2)$$

Are you sure about the tan substitution? I couldn't solve it that way, but I'm not very good

 

[Saitama]

Are you sure about the tan substitution? I couldn't solve it that way, but I'm not very good

Hi dumbperson! :)

Yes, I am sure about it. After the substitution, you should get:
$$I=\int_0^{\pi/4} \ln(1+\tan\theta)\,d\theta$$
Also,
$$I=\int_0^{\pi/4} \ln\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)\,d\theta=\int_0^{\pi/4} \ln\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)\,d\theta=\int_0^{\pi/4} \ln(2)-\ln(1+\tan\theta) \,d\theta$$
$$\Rightarrow I=\frac{\pi \ln2}{4}-I \Rightarrow \boxed{I=\frac{\pi \ln2}{8}}$$

I hope that helps.

 

[dumbperson]

Hi dumbperson! :)

Yes, I am sure about it. After the substitution, you should get:
$$I=\int_0^{\pi/4} \ln(1+\tan\theta)\,d\theta$$
Also,
$$I=\int_0^{\pi/4} \ln\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)\,d\theta=\int_0^{\pi/4} \ln\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)\,d\theta=\int_0^{\pi/4} \ln(2)-\ln(1+\tan\theta) \,d\theta$$
$$\Rightarrow I=\frac{\pi \ln2}{4}-I \Rightarrow \boxed{I=\frac{\pi \ln2}{8}}$$

I hope that helps.

Ah cool, thanks!

 

[acegikmoqsuwy]

Not really a hard one but try this: \displaystyle\int 1-\sin x\cos x\tan x\,dx

 

[STEMucator]

Some of my most memorable ones from calc I, should take a few coffees to solve:

$$\int_{0}^{∞} \frac{1}{x^2 + \sqrt{x}} dx$$

$$\int_{0}^{\frac{\pi}{4}} \frac{1}{1 + sin^2(x)} dx$$

 

[exo]

$$\int_{0}^{\frac{\pi}{4}} \frac{1}{1 + sin^2(x)} dx$$

if we divide both the numerator and the denominator by \frac{1}{cos^{2}x} the integral becomes

\displaystyle\int_{0}^{\frac{\pi}{4}} \frac{sec^{2}x}{1+2tan^{2}x} dx

We can make the substitution t = tan(x)

I guess you can solve the first one by making the sub u = \sqrt{x} and then using partial fractions decomposition

 

[einstein314]

This one was a 1968 Putnam competition problem, I believe:
{\int_{0}^{1}{\frac{x^4 (1-x)^4}{1+x^2} dx}}
The answer is really interesting...
If you're really up for a challenge, try this continuation:
{\int_{0}^{1}{\frac{x^8 (1-x)^8 (25+816x^2)}{3164(1+x^2)} dx}}
These are just tedious and definitely easier than many of the problems here.
-- Joseph

 

[dextercioby]

The 1st one is not tedious:

- \int_{0}^{1} \frac{(1-x^4 -1)(1-x)^4}{1+x^2} dx = - \int_{0}^{1} (1-x)^5 (1+x) dx + \int_{0}^{1} \frac{(1-x)^4}{1+x^2} dx

For the 1st integral, just sub 1-x = p and it will be trivial.

For the 2nd integral, write the integrand as

\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.

 

[GFauxPas]

The 1st one is not tedious:

- \int_{0}^{1} \frac{(1-x^4 -1)(1-x)^4}{1+x^2} dx = - \int_{0}^{1} (1-x)^5 (1+x) dx + \int_{0}^{1} \frac{(1-x)^4}{1+x^2} dx

Can you explain in more detail exactly what you did there please? I see that $x^4 = -(-x^4 +1 - 1)$ and then I don't see how you got to the next step.

I had a good time with this one:

\int_{\to 0}^{\to 1}\ln x \ln (1-x) \, \mathrm dx

(+7 points if you prove that it exists before evaluating it)

 

[Saitama]

I had a good time with this one:

\int_{\to 0}^{\to 1}\ln x \ln (1-x) \, \mathrm dx

(+7 points if you prove that it exists before evaluating it)

$$\int_0^1 \ln x\ln(1-x)\,dx=-\sum_{k=1}^{\infty} \frac{1}{k}\int_0^1 x^k\ln x\,dx=\sum_{k=1}^{\infty} \frac{1}{k(k+1)^2}$$
$$=\sum_{k=1}^{\infty} \frac{1+k-k}{k(k+1)^2}=\sum_{k=1}^{\infty} \frac{1}{k(k+1)}-\sum_{k=1}^{\infty} \frac{1}{(k+1)^2}$$
Both the sums are easy to evaluate, the first one evaluates to 1 by telescoping series and the second one is simply $\zeta(2)-1$, hence, the final answer is $2-\zeta(2)$.

This one was a 1968 Putnam competition problem, I believe:
{\int_{0}^{1}{\frac{x^4 (1-x)^4}{1+x^2} dx}}
The answer is really interesting...
If you're really up for a challenge, try this continuation:
{\int_{0}^{1}{\frac{x^8 (1-x)^8 (25+816x^2)}{3164(1+x^2)} dx}}
These are just tedious and definitely easier than many of the problems here.
-- Joseph

http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_π

 

[dextercioby]

Ha, nice trick to find an article on wikipedia about those 2 integrals. :)

 

[guysensei1]

Not terribly hard but something I've found interesting
\int_{0}^{1}\sin (\ln(x)) dx

 

[Curious3141]

Not terribly hard but something I've found interesting
\int_{0}^{1}\sin (\ln(x)) dx

This one is really easy. Sub x=e^y. The rest is either that trick with the repeated integration by parts or transforming \sin y to a complex exponential form.

 

[zaidalyafey]

$$\int^1_0 \log^2(x)\log^2(1-x)\,dx$$

 

[phion]

\int^2_0 \sin{x^2}\,dx

 

[dextercioby]

\int^2_0 \sin{x^2}\,dx

There's nothing hard about it and nothing interesting either, as long you've read about special functions like http://mathworld.wolfram.com/FresnelIntegrals.html

 

[dextercioby]

Compute:

$$\int_{0}^{1}\int_{0}^{1} \frac{y^2 -x^2}{(x^2 +y^2)^2} dx{}~{}dy $$

 

[phion]

There's nothing hard about it and nothing interesting either, as long you've read about special functions like http://mathworld.wolfram.com/FresnelIntegrals.html

Solve it then? Approximation techniques don't count either.

 

[h6ss]

I'm curious to see what technique you guys would use to solve $\int_{-1}^{1} \frac{\cos(x)}{\exp(1/x) + 1}dx$

 

[quantumtimeleap]

\int e^{-x^2} dx

how can this even be integrated?

You don't. You use a power series for that one. Hahaha.

 

[Calculus Master]

sin(2x)cos(2x)dx

\int{\sin{2x}\cos{2x}dx} = \frac{1}{2} \int{\sin{4x}} = -\frac{1}{8} \cos{4x}

 

[Calculus Master]

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

To integrate \int_{0}^{\infty}sin(x^2)dx we use \int_{0}^{\infty}sin(x^n)dx =\Gamma{(1 + \frac{1}{n})}\sin{\frac{\pi}{2n}} = \frac{\sqrt{\pi}}{2\sqrt{2}}

 

[STEMucator]

Give this one a chug:

$$\int x \sqrt{ax^2 + bx + c}$$

 

[Curious3141]

Give this one a chug:

$$\int x \sqrt{ax^2 + bx + c}$$

I would rewrite it as $\displaystyle \frac{1}{2a}[\int(2ax + b)\sqrt{ax^2 + bx + c}dx - b\int\sqrt{ax^2 + bx + c}dx]$

The first integral is of the form $\displaystyle \int f'(x)g(f(x))dx$ and can be solved easily by either inspection or applying the sub$\displaystyle ax^2 + bx + c = u$.

The second integral is evaluated by completing the square and applying a hyperbolic sine sub.

 

[hEMU]

Disclaimer: A friend told me about this and I do not know the answer yet.
Disclaimer 2: This is not really an integration problem but very close. (If you think it should not be here, please tell me, I will remove it)

Try this one,
(dy/dx) + 3 = (dy/dx) + 2
Apparently, it does have a solution (or so I am told and yes, I am still working on the problem)

 

[zoki85]

Disclaimer: A friend told me about this and I do not know the answer yet.
Disclaimer 2: This is not really an integration problem but very close. (If you think it should not be here, please tell me, I will remove it)

Try this one,
(dy/dx) + 3 = (dy/dx) + 2
Apparently, it does have a solution (or so I am told and yes, I am still working on the problem)

Any vertical line (x=a) satisfies.
"∞ = ∞"

 

[acegikmoqsuwy]

\displaystyle\int_0^\infty e^{-\sqrt[2015] x} dx = 2015!

True or false? Prove your answer. Came up with this one myself :D

 

[dextercioby]

That's true, I think. I'm lazy to do the maths, but you can make an obvious substitution, then use the definition of the Euler Gamma function.

 

[cpman]

This one isn't too hard, but it is really cool:
\int^{\infty}_{-\infty} \frac{1}{x^2 + 1} \, \mathrm{d}x

 

[hEMU]

This one isn't too hard, but it is really cool:
\int^{\infty}_{-\infty} \frac{1}{x^2 + 1} \, \mathrm{d}x

Wouldn't it just be zero?
-You substitute x = tan u
-You get [arctan x] from (-inf) to (+inf) which is 0