- #71
jhae2.718
Gold Member
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[tex]\int\!\!\sqrt{x^2+1}\,\mathrm{d}x[/tex] is a fun one...
andyb177 said:(tanx)^1/2
gb7nash said:[tex]\int\!\!\sqrt{x^2+1}\,\mathrm{d}x[/tex] 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 said: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
[tex] \sin x = \mbox{Im}\left(e^{ix}\right) [/tex].
chandi2398 said:see the question_attached and please let me know your method to solve this. :)
This one is fairly straightforward, use the substitution ##x=\tan\theta##.dumbperson said:[tex] \int^1_0 \frac{\ln(1+x)}{1+x^2} dx[/tex]
For this one, define:[tex]\int^1_0 \frac{x-1}{\ln(x)} dx [/tex]
We use the series expansion of ##\ln(1-x)## i.e[tex] \int^1_0 \frac{\ln(1-x)}{x} dx[/tex]
Goa'uld said:Here is an interesting one I came up with.
[tex]\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\log(\cos(x))dx}{e^x+1}[/tex]
Pranav-Arora said: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:
$$-\int_0^1 \frac{1}{x}\sum_{k=1}^{\infty} \frac{x^k}{k}\,dx$$
$$=-\sum_{k=1}^{\infty} \int_0^1 \frac{x^{k-1}}{k}\,dx$$
$$=-\sum_{k=1}^{\infty} \frac{1}{k^2}=-\zeta(2)$$
dumbperson said:Are you sure about the tan substitution? I couldn't solve it that way, but I'm not very good
Pranav-Arora said: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.
Zondrina said:$$\int_{0}^{\frac{\pi}{4}} \frac{1}{1 + sin^2(x)} dx$$
dextercioby said:The 1st one is not tedious:
[tex] - \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 [/tex]
GFauxPas said:I had a good time with this one:
[tex]\int_{\to 0}^{\to 1}\ln x \ln (1-x) \, \mathrm dx [/tex]
(+7 points if you prove that it exists before evaluating it)
einstein314 said:This one was a 1968 Putnam competition problem, I believe:
[tex]{\int_{0}^{1}{\frac{x^4 (1-x)^4}{1+x^2} dx}}[/tex]
The answer is really interesting...
If you're really up for a challenge, try this continuation:
[tex]{\int_{0}^{1}{\frac{x^8 (1-x)^8 (25+816x^2)}{3164(1+x^2)} dx}}[/tex]
These are just tedious and definitely easier than many of the problems here.
-- Joseph
This one is really easy. Sub [itex]x=e^y[/itex]. The rest is either that trick with the repeated integration by parts or transforming [itex]\sin y[/itex] to a complex exponential form.guysensei1 said:Not terribly hard but something I've found interesting
[tex]\int_{0}^{1}\sin (\ln(x)) dx[/tex]