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

[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

 

[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
\[ \int \frac{ 2u^2}{1+ u^4} du = \int \frac{ u^2 + 1}{1+u^4} du + \int \frac{ u^2 - 1}{1+u^4} du \]<br /> <br /> \[ = \int \frac{ 1+ \frac{1}{u^2} }{u^2 + \frac{1}{u^2} } du + \int \frac{ 1- \frac{1}{u^2} }{u^2 + \frac{1}{u^2} } du \]<br /> <br /> \[ = \int \frac{ d\left( u - \frac{1}{u} \right) }{ \left( u - \frac{1}{u} \right)^2 +2 } + \int \frac{ d\left( u + \frac{1}{u} \right) }{ \left( u + \frac{1}{u} \right)^2 -2 } \]<br /> <br /> \[ = \frac{ 1}{\sqrt{2}} \left( \tan^{-1} \left( \frac{ u - \frac{1}{u}}{\sqrt{2}} \right) - \tanh^{-1} \left( \frac{ u + \frac{1}{u}}{\sqrt{2}} \right) \right) + C \]<br /> <br /> \[ = \frac{ 1}{\sqrt{2}} \left( \tan^{-1} \left( \frac{ \sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}} \right) - \tanh^{-1} \left( \frac{ \sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{2}} \right) \right) + C \]

 

[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.