Is it true that $\int_{- \infty }^{\infty} \frac{1}{1+x^2}\, dx=[\tan^{-1} x]_{- \infty }^{\infty} = \frac{\pi}{2}-(-\frac{\pi}{2})=\pi$?  I'm integrating after a long period of time of inactivity, and sometimes I have doubts.
122196
How $$\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$$ exists?   It is difficult question to me. i have tried to evaluate by using fact that $$\int_{-\infty}^{\infty} f(x) \ dx =\int_{-\infty}^{0} f(x)\, dx + \int_{0}^{\infty} f(x)\, dx$$ but i have failed in this one.   any hints on this one?
102137
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$   My approach :   Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then  $$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$  But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
109204
I want to show that  $$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2\sqrt{a(a+1)}}$$  for $a &gt; 0$.  I try several methods   Substitutions to rationalize The famous $U = \tan(\frac{\theta}{2})$ Multiplying by conjugates   and other calculus techniques and still, I can not prove the equality.   Anyone can give me a hint of how should I begin to work this integral?  I'm very ashamed that I can not solve this problem.
56865
Evaluate the following integral.$$\int \frac{\sec x}{\sqrt{3+\tan x}}dx$$  On putting $t=\tan x$ I am getting $$\int\frac{1}{\sqrt{(t^2+1)(t+3)}}dt$$.  How should i proceed from here.
110711
$$\int_{-\infty}^{\infty}{{\rm d}x \over 1 + x^{2n}}$$  How to calculate this integral? I guess I need to use residue. But I looked at its solution. But it seems too complicated to me. Thus, I asked here. Thank you for help.
19082
Compute the integral.   $$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$  The answer at the back of the book is  $$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$
46297
Find $$M:=\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$$   There's a solution  that uses complex numbers which I didn't understand and I was wondering if the following is also a correct method.   My proposed solution   $$\begin{align} &amp;\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&amp;\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =&amp;\sum_{n=1}^{\infty}(\tan^{-1}(1+n)+\tan^{-1}(1-n))\\ =&amp;\sum_{n=1}^{\infty}(\tan^{-1}(n+1)-\tan^{-1}(n-1)) \end{align} $$  And this implies  $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\frac{3\pi}{4}$$
86824
Could someone point me to an online resource where I can look up  $$ \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\cos(\vec k\cdot\vec x)}{\sqrt{|\vec k|^2+m^2}}\,dk_1\,dk_2\,dk_3$$  where $\vec k$ and $\vec x$ are 3-vectors, and $m$ is a constant.  Also, if you know how to do it step by step and want to show off your fancy integration, that would also be very helpful.  Thanks ;)
3970
I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$ What do you think?  It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this .
48681
$$\frac {2\tan {x}}{1-(\tan {x})^2} = (\sin {2x})^2$$  I tried a lot but I get nowhere
52543
I have no idea how to start, it looks like integration by parts won't work.  $$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$  If someone could shed some light on this I'd be very thankful.
116454
Given that $\int_{-\infty}^\infty e^{-x^2} \, dx=\sqrt{\pi}$. Find the value of $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+xy+y^2)} \, dx\,dy$$  I don't understand how I find this double integral by using the given data. Please help.
70652
Numerically it seems to be true that  $$ \int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}. $$  Any ideas how to prove this?
75155
So I came across this question on brilliant.org and didn't know how to go about it: $$\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $$  I tried to complete the squares in the denominator and then use a trigonometric substitution (with tan) but I didn't get anything beyond that.   How do I do it? Do we need to differentiate under the integral sign here?
54941
$$ \int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2) $$  Anyone an idea on how to prove this?
87933
$$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$  My Attempt:  $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$  Replacing $x$ by $1-x$,we get  $$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$  Then I got stuck. Please help.
47463
It is quite easy to show that (by using residue theorem) $$\int_0^{\infty} \frac{1}{1+x^n}dx = \frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)} $$ for $$n \ge 2$$  Is there any possibility to simplify $$\frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)}$$ or it is best result?  Thanks in advance!