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Since no one has replied, I will give a go. I initially set out trying this with contours. It can be done, but I do not have the gumption to endeavor. It is rather involved. $$2\int_{0}^{\frac{\pi}{2}}\frac{x\cos(x)}{1+\sin^{2}(x)}dx$$ Using parts let $$u=x, \;\ dv=\frac{\cos(x)}{1+\sin^{2}(x)}dx$$ ...

Hey Fellas: Kudoes as always on the ingenious solutions. Sorry for not being around much, but work takes up my time through the week. We already have several problems to catch up on. A week or two ago I posted a series, then deleted it. Well, here it is again. I think it is a cool one. I post more a...

No one wanted to try this one?. Well, the reason I have not posted my solution is merely laziness. I think this is a fun way, though more than likely not the easiest or even best way. I think it is interesting, nonetheless. Write as $$2\int_{\pi/4-a}^{\pi/4+a}xcsc(2x)dx$$ Let $u=2x$ and obtain: $$1/...

Show that:

$$\sum_{n=1}^{\infty}\left(2n\log\left(\frac{4n+1}{4n-1}\right)-1\right)=\frac{1}{2}-\frac{1}{\pi}G-\frac{1}{4}\log(2)$$

G is a priori....the Catalan

There...some math Latin

You sure made short work of that one, T. Wow!

Go RD

Here's a kick on a famous one:

$$\int_{0}^{1}\frac{(\sin^{-1}(x^{2}))^{2}}{\sqrt{1-x^{2}}}dx=\frac{\pi^{3}}{4}-\frac{3\pi}{4}\log^{2}(2)-2\pi Li_{2}\left(\frac{1}{\sqrt{2}}\right)$$

I like to use the relation $$\int_{a}^{b}p(x)\cot(x)dx=2\sum_{k=1}^{\infty}\int_{a}^{b}p(x)\sin(2kx)dx$$ when doing integrals that involve the product of a polynomial and cot There are also similar relations for csc and so forth. If this were csc instead of cot, we would use $\sin[(2k+1)x]$ instead....

$$\int_{0}^{1}\frac{\ln(1+x+x^{2}+\cdot\cdot\cdot +x^{n-1})}{x}dx$$

Show that:

$$\int_{\frac{\pi}{4}-a}^{\frac{\pi}{4}+a}\frac{x}{\sin(x)\cos(x)}dx=\pi \tanh^{-1}(\tan(a))$$