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

[jhae2.718]

$$\int\!\!\sqrt{x^2+1}\,\mathrm{d}x$$ is a fun one...

 

[gb7nash]

$$\int\!\!\sqrt{x^2+1}\,\mathrm{d}x$$ looks like a problem I'd be banging my head against the wall to solve. I'm assuming you use tan² + 1 = sec² and some funny manipulation?

 

[dextercioby]

Ok, people are accustomed to solve the $\int \sqrt{x^2 +1} \, dx$ by some trigonometric substitution, either hyperbolic sine, or circular secant/cosecant.

But there's a third way which is not transcendental until the end. The substitution

$$\sqrt{x^2 +1} - x = t$$

 

[andyb177]

(tanx)^1/2

 

[dextercioby]

(tanx)^1/2

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

is a well-known elliptic integral. It can be evaluated by Mathematica explicitely.

 

[icesalmon]

$$\int\!\!\sqrt{x^2+1}\,\mathrm{d}x$$ looks like a problem I'd be banging my head against the wall to solve. I'm assuming you use tan² + 1 = sec² and some funny manipulation?

Why couldn't you just perform a general u-substitution with this setting u = x^2 + 1 du =2xdx, dx= du/2x -> int [sqrt(u)/2x]du x=+/-sqrt(u-1)
lol, nevermind. Sorry!

 

[Gib Z]

If we let $$u^2 = \tan x$$ we have
$$\[ \int \frac{ 2u^2}{1+ u^4} du = \int \frac{ u^2 + 1}{1+u^4} du + \int \frac{ u^2 - 1}{1+u^4} du \] \[ = \int \frac{ 1+ \frac{1}{u^2} }{u^2 + \frac{1}{u^2} } du + \int \frac{ 1- \frac{1}{u^2} }{u^2 + \frac{1}{u^2} } du \] \[ = \int \frac{ d\left( u - \frac{1}{u} \right) }{ \left( u - \frac{1}{u} \right)^2 +2 } + \int \frac{ d\left( u + \frac{1}{u} \right) }{ \left( u + \frac{1}{u} \right)^2 -2 } \] \[ = \frac{ 1}{\sqrt{2}} \left( \tan^{-1} \left( \frac{ u - \frac{1}{u}}{\sqrt{2}} \right) - \tanh^{-1} \left( \frac{ u + \frac{1}{u}}{\sqrt{2}} \right) \right) + C \] \[ = \frac{ 1}{\sqrt{2}} \left( \tan^{-1} \left( \frac{ \sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}} \right) - \tanh^{-1} \left( \frac{ \sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{2}} \right) \right) + C \]$$

 

[Nebuchadnezza]

Would anyone be interested if I started a new topic here, and posted all the integrals I have ?

I have spent perhaps a week going through my book and various webpages to find all the challenging and interesting integrals I could. Some are taken from here but many are not.

I think I have about 100-120 integrals. Ranging from easy to really hard. Would anyone be interested in that? Ofcourse I could post all of them here, but it would be messy. Much nicer with a first post containing integrals.

 

[snipez90]

Sure why not. I'll even attempt to do them using complex analysis. Maybe.

 

[Q U A N T U M]

$\int xe^{ax}sin(bx) dx$

I LOVED solving this one; took an intellectual pleasure in it, to be honest :P .

It's not hard actually, just quite a bit of work.

 

[HomogenousCow]

you guys really love your taylor expansions

 

[chandi2398]

Check this



∫1/(1+x⁴) dx

 

[Goa'uld]

Here is an interesting one I came up with.
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\log(\cos(x))dx}{e^x+1}$$

 

[chandi2398]

Please give me an answer!

see the question_attached and please let me know your method to solve this. :)

 

[dextercioby]

First you can drop the absolute value for the sine, since it's completely positive on the integration domain. Then I'd use a trick writing

$$\sin x = \mbox{Im}\left(e^{ix}\right)$$.

 

[Goa'uld]

First you can drop the absolute value for the sine, since it's completely positive on the integration domain. Then I'd use a trick writing

$$\sin x = \mbox{Im}\left(e^{ix}\right)$$.

That method doesn't work since it arrives at a different result than the correct answer. The imaginary part cannot be taken after the integral to obtain the same answer since the logarithm doesn't work that way.

 

[Curious3141]

see the question_attached and please let me know your method to solve this. :)

The correct way to evaluate this integral is as follows. Call the integral $\displaystyle I$.

Sub $\displaystyle y = \frac{\pi}{2} - x$.

Now you can prove that $\displaystyle I = \int_0^\frac{\pi}{2} \ln \cos y dy = \int_0^\frac{\pi}{2} \ln \cos x dx$.

Hence $\displaystyle 2I = \int_0^\frac{\pi}{2} \ln (\frac{1}{2}\sin 2x) dx = \int_0^\frac{\pi}{2}(-\ln 2)dx + \int_0^\frac{\pi}{2}\ln \sin 2x dx = -\frac{\pi\ln 2}{2} + \int_0^\frac{\pi}{2}\ln \sin 2x dx$.

Now $\displaystyle \int_0^\frac{\pi}{2}\ln \sin 2x dx = \frac{1}{2}\int_0^{\pi}\ln \sin x dx$ as should become apparent after another sub of $\displaystyle u = 2x$.

$\displaystyle \frac{1}{2}\int_0^{\pi}\ln \sin x dx = (2)(\frac{1}{2})\int_0^{\frac{\pi}{2}}\ln \sin x dx = I$ by symmetry.

So you're left with $\displaystyle 2I = I -\frac{\pi\ln 2}{2}$ yielding $\displaystyle I = -\frac{\pi\ln 2}{2}$

 

[dumbperson]

3 very hard ones (spoiler alert: they do not have an anti derivative)

$$\int^1_0 \frac{\ln(1+x)}{1+x^2} dx$$


$$\int^1_0 \frac{x-1}{\ln(x)} dx$$

$$\int^1_0 \frac{\ln(1-x)}{x} dx$$

 

[Saitama]

$$\int^1_0 \frac{\ln(1+x)}{1+x^2} dx$$

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

$$\int^1_0 \frac{x-1}{\ln(x)} dx$$

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,
$$I(1)=\ln(2)$$

$$\int^1_0 \frac{\ln(1-x)}{x} dx$$

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:
$$-\int_0^1 \frac{1}{x}\sum_{k=1}^{\infty} \frac{x^k}{k}\,dx$$
$$=-\sum_{k=1}^{\infty} \int_0^1 \frac{x^{k-1}}{k}\,dx$$
$$=-\sum_{k=1}^{\infty} \frac{1}{k^2}=-\zeta(2)$$

 

[Saitama]

Here is an interesting one I came up with.
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\log(\cos(x))dx}{e^x+1}$$

Let
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\log(\cos(x))dx}{e^x+1}$$
We can also write:
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\log(\cos(x))dx}{e^{-x}+1}$$
Add both the expressions for I to get:
$$2I=\int_{-\pi/2}^{\pi/2} ln(\cos(x))\,dx=2\int_0^{\pi/2}ln(\cos(x))\,dx$$
I can rewrite the above as:
$$I=\int_0^{\pi/2} \ln(\sin(x))$$
The above definite integral is evaluated by Curious3141 in his post #87.

 

[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]

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