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Re: MX2 2015 Integration Marathon $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\cot x}{\cot^2 x+\sqrt{\cot x}}dx$

Re: MX2 2015 Integration Marathon $Let \displaystyle I = \int\frac{1}{\sin^3 x+\cos^3 x}dx = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x)}dx$ $So we can write it as \displaystyle \frac{1}{3}\int \left[\frac{2}{\sin x+\cos x}+\frac{(\sin x+\cos x)}{1-\sin x\cdot \cos x}\right]dx$...

Re: MX2 2015 Integration Marathon $Let \displaystyle I = \int\frac{\sin^2 x}{\cos^2 x+\cot^2 x}dx = \int\frac{\sin^4 x}{(1+\sin^2 x)\cdot \cos^2 x}dx = \int\frac{(\sin^4 x-1)+1}{(1+\sin^2 x)\cdot \cos^2 x}dx$ $So Integral \displaystyle I = -\int 1\cdot dx+\int\frac{1}{(1+\sin^2 x)\cdot...

Re: MX2 2015 Integration Marathon $ Evaluation of $\displaystyle \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx$

Re: MX2 2015 Integration Marathon $Let\; $I = \sqrt{\tan x}\;\mathrm{d}x$\; and \; $J = \sqrt{\cot x}\;\mathrm{d}x$. $\displaystyle I + J = \int\left(\sqrt{\tan x} + \sqrt{\cot x}\right) \;\mathrm{d}x$ $\displaystyle = \sqrt{2} \int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \;\mathrm{d}x$...

Re: MX2 2015 Integration Marathon $Let \; $\displaystyle \mathcal{I} = \int_{-1}^{1}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{1+2^{\frac{1}{x}}} \right )\text{ d}x$ = \displaystyle \left[\frac{1}{1+2^{\frac{1}{x}}}\right]_{-1}^{1} = \displaystyle...

Re: MX2 2015 Integration Marathon $Let $\displaystyle \mathcal{I} = \frac{1}{x^7+x}dx = \int\frac{x^{-7}}{1+x^{-6}}dx$ $Now \; let $(1+x^{-6})dx = t\;,$ Then \; $\displaystyle x^{-7}dx = -\frac{1}{6}dt$ $So\; Integral $\displaystyle I = -\frac{1}{6}\int \frac{1}{t}dt = -\frac{1}{6}\ln...

Re: MX2 2015 Integration Marathon $Evaluation\; of\; $\displaystyle \int \frac{4x^5-1}{(x^5+x+1)^2}dx$\; and $\displaystyle \int \sqrt{1+\tan x \cdot \tan (x+\alpha)}dx$

Re: MX2 Integration Marathon $Calculation \; of \; Integral \; $\displaystyle \int\frac{\sin^2(nx)}{\sin^2 (x)}dx$\;, where \; $n\in \mathbb{N}$

$Using\; Combinatorial \; argument \; How\; can\; we\; prove \; $ \displaystyle \frac{n^2!}{(n!)^2}$\; is \; an $ Integer \; Quantity\;,where \; $n\in \mathbb{N}$

Re: MX2 Integration Marathon $Calculate\; the \; Integrals $(a)\;\; \displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n}(x)\cdot \sin (nx)dx$\;\;\;\;\;\;(b)\;\; \int_{0}^{\frac{\pi}{2}}\cos^{n}(x)\cdot \cos (nx)dx$ $Where \; $n\in \mathbb{N}$

Re: MX2 Integration Marathon $Let \; $\displaystyle A = \sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}$\; and $\displaystyle B = \sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2}.$ $for \;$n=1,2,3,4,..........,$\; Then\; which \; one \; is \; Leargest...$

Re: MX2 Integration Marathon $Assuming\; $\alpha >0$ \;,Now\; let\;\displaystyle I = \int_{\frac{1}{\alpha}}^{\alpha} \frac{\tan^{-1} x}{x}dx$ $Now\; let \; $\displaystyle x = \frac{1}{t}$\;,Then\; $\displaystyle dx = -\frac{1}{t^2}$\; and \; changing \; limits\; we \; get$ $Now \; let \...

Thanks Square3root Using Yours Hint. $Using Integration by parts for second $ $Let $\displaystyle I = \int_{\frac{\pi}{2}}^{\pi}x\cdot \cot xdx = \left[x\cdot \ln(\sin x)\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\ln (\sin x)dx$ Now How can I solve after that

$Evaluation of $\displaystyle \int_{0}^{\pi}x\cdot \cot x dx$ $I have Tried like this way.$ $Here $\displaystyle \cot x$ is $0$at $\displaystyle x = \frac{\pi}{2}$. $So we can break integral$ $ as $\displaystyle \int_{0}^{\frac{\pi}{2}}x\cdot \cot xdx +\int_{\frac{\pi}{2}}^{\pi}x\cdot \cot x...

$\displaystyle I = \int_{0}^{2}\left(3x^2-3x+1\right)\cdot \cos\left(x^3-3x^2+4x-2\right)dx$

$Evaluation\; of \; sum \; $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{n^2}\sum_{k=1}^{n}\sqrt{n^2-k^2}$

Re: MX2 Integration Marathon $I\; have \ tried \; like \; this \; way....$ $\displaystyle \int \tan x\cdot \tan 2x\cdot \tan 3xdx = \int \tan x \cdot \frac{2\tan x}{1-\tan^2 x}\cdot \frac{3\tan x-\tan^3 x}{1-3\tan^2 x}dx$ $Now\; Let \; $\tan x= t\;,$ Then \; $\displaystyle \sec^2 xdx =...

Re: MX2 Integration Marathon $\displaystyle \int\frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}dx$ $\displaystyle \int x^{26}\cdot \left(x-1\right)^{17}\cdot (5x-3)dx$

Re: MX2 Integration Marathon $\displaystyle I = \int \frac{1}{\sqrt{x}\cdot \left(\sqrt[4]{x}+1\right)^{10}}dx$ $\displaystyle J = \int\frac{1+x^2}{\left(1-x^2\right)\sqrt{1+x^4}}dx$ $\displaystyle K = \int\frac{1}{\left(1-x^2\right)\sqrt{1+x^4}}dx$ $\displaystyle L = \int \frac{\cos^2...